CHEMISTRY COMMUNITY

Nada AbouHaiba 1I

Usually a metal in the solid phase is what you want to look for

Nada AbouHaiba 1I

Usually it will be explicitly stated, but if not you can tell which is the slow/fast step by looking at other information given.
For example, if a energy profile is given you want to pick the one with the lowest activation energy and that would be the fast step relative to the other.

Nada AbouHaiba 1I

You would give the molecularity of each step separately or if asked for the molecularity of the overall reaction you would go ahead and write the overall rxn from steps 1 and 2 and the write the molecularity of that overall rxn. You calculate molecularity by counting up how many moles are in the rea...

Nada AbouHaiba 1I

This has to do with homework problem 15.95 in the 6th edition For part of e of this question it says that a catalyst is speeding up the third reaction only and then asks what this does for the rate of the overall reaction. The answer is that there is no change and I was just a little confused to as ...

Nada AbouHaiba 1I

The problem tells you that N2O5 decomposes by a first order reaction, so the rate law will have the form rate = k[A] where A is the reactant, N2O5, and the it is raised to the first power since it is a first order reaction. You are also given k = 5.2 x 10^-3. As mentioned before, you don't need to ...

Nada AbouHaiba 1I

the m and n powers do not have units, they are just experimentally determined values

Nada AbouHaiba 1I

The initial concentration just cancel each other out when the half life equation is being derived

Nada AbouHaiba 1I

Trans has the higher entropy because it is able to occupy more possible positions.

Nada AbouHaiba 1I

I know you want to look at the most negative reduction potential in order to find which is the strongest reducing agent, but how do you know which equation to use if there are several that have that same metal in the back? 14.25 Arrange the following metals in order of increasing strength as reducin...

Nada AbouHaiba 1I

To see which way the reaction proceeds you would just compare Q and K with one another. If Q is less than K then it will proceed to the right and if Q is greater than K then it will proceed left.

Nada AbouHaiba 1I

Gibbs Free Energy will also be on the test! I believe Lavelle posted an announcement on his website on what would be on the test

Nada AbouHaiba 1I

You're going to want to identify what was reduced and what was oxidized by looking at the different oxidation states. After that you want to balance the reactions and cancel!

Nada AbouHaiba 1I

That's considered oxidation! The way I like to tbhink about it is if the charge was REDUCED then it was REDUCTION

Nada AbouHaiba 1I

Acidic reactions give off electrons whereas basic reactions gain electrons

Nada AbouHaiba 1I

I believe galvanic cells are just units that turn the chemical energy from redox reactions into electrical energy.

Nada AbouHaiba 1I

Here's an article I found about reduction potentials!
https://socratic.org/questions/how-do-y ... -potential

Nada AbouHaiba 1I

I don't have a list but I have two assumptions :
1. When a rxn is isothermal/reversible delta S Universe = 0
2. When a rxn is irreversible delta S Surroundings = 0

Nada AbouHaiba 1I

Here's a link!
https://lavelle.chem.ucla.edu/wp-conten ... ations.pdf

Nada AbouHaiba 1I

I guess what it comes down to is if you have enough information to calculate moles using pv=nrt, so if you're given all the information then you'd solve for n.

Nada AbouHaiba 1I

Ronald Thompson 1F wrote:Is there a solutions guide as well?

On the post Lyndon said that we shouldn't expect a solutions guide, so I'm assuming the only way to get answers is to go to the review session or get them from a buddy

Nada AbouHaiba 1I

So for this problem you're gonna wanna use the ideal gas equations, Cp (constant pressure) is 5/2 of the gas constant (which is 8.314) and Cv (constant volume) 3/2 of the same gas constant. One you calculate the C for each scenario you'll plug it into equation nCdeltaT to solve for q which is the he...

Nada AbouHaiba 1I

I don't think so just because delta U is made up by the sum of heat and work done and work can still be done even if there's constant temperature.

Nada AbouHaiba 1I

I think Lavelle said something about what we learn up to Entropy will be the only thing we're tested on from Unit 4

Nada AbouHaiba 1I

LedaKnowles2E wrote:What's delta U?

Delta U is the change in internal energy

Nada AbouHaiba 1I

They're just given as positive values, but if you're forming the bond then you would write them as negative because it is an exothermic process

Nada AbouHaiba 1I

Megan_Ervin_1F wrote:Why does this section talk about delta U? I don't recall learning about delta U in lecture...

I think we haven't learned about it yet. Lavelle in class today mentioned how he is teaching the material in a different order than the textbook goes.

Nada AbouHaiba 1I

It mentioned in the textbook that if the second protonation is less than 1/1000 of the original one you can just solve for the first one and use that value of H3O/OH to calculate pH/pOH. It's because the second protonation would be so small there it would make no difference. H2SO4 is the exception i...

Nada AbouHaiba 1I

Here's a list of the strong acids and bases! You can assume everything else is weak. Hope this helps!

Nada AbouHaiba 1I

I don't believe there's any equation that will relate the two. You would just have to use an ICE box, solve for the concentration of H3O/OH and then calculate pH/pOH

Nada AbouHaiba 1I

Here's a picture of the strong acids and bases!

Nada AbouHaiba 1I

When using ICE tables, how do you know when its better to use the quadratic formula to find X or the 5% rule? If you're given a weak acid or base with an equilibrium constant less than 10^-3 you can assume that the x in the (number-x) is negligible and you can just get rid of that x. If it's not a ...

Nada AbouHaiba 1I

Yes! Professor Lavelle mentioned that anytime the K value is less than 10^-3, the approximation of x in the reactant-x is negligible.

Nada AbouHaiba 1I

To find the equilibrium constant you'll just use the equation concentration of the products/concentrations of the reactants, not pv=nrt.

Nada AbouHaiba 1I

Usually you can tell by looking at whether or not the reaction is overall endothermic or exothermic. Usually with problems like that they should give you a delta H or just tell you.

Nada AbouHaiba 1I

Yes, I believe so

Nada AbouHaiba 1I

I don't think that's been announced because I believe that's still being decided by Lavelle.
Your best bet for getting a solid answer on this might be emailing Lavelle (also if he does tell you this info plz share with the rest of chemistry community)

Nada AbouHaiba 1I

Intramolecular forces are those that bind elements together, so for example covalent and ionic bonds.
Intermolecular forces are those that bind molecules or whole compounds together, so LDFs or H-Bonds.

Nada AbouHaiba 1I

Yes! All you need to know regarding shrodinger's equation is that it was what was used to calculate the electron densities in a cell and allowed for the "creation" of the orbitals!

Nada AbouHaiba 1I

Yes, polarity does depend on whether the molecule is cis or trans.
I don't believe we will ever be given a molecule that could be either and asked about it's polarity if it's not specifically told which way it should be drawn.

Nada AbouHaiba 1I

The first molecule possesses iodine atoms, which are substantially larger than the fluorine atoms found in the second molecule. With a larger atomic number, iodine has more electrons and a larger electron cloud. Iodine, therefore, has a larger polarizability value than fluorine and forces a group o...

Nada AbouHaiba 1I

Can someone explain why CHI4 has a higher melting point than CHF4. Thanks!

Nada AbouHaiba 1I

https://www.youtube.com/watch?v=ZnP5vavICZw
this is a different way to remember significant figures
it's helpful if you get confused between decimal present and no decimal

Nada AbouHaiba 1I

I don't think we'll need to know how to draw the d orbital because it's complex compared to s and p

Nada AbouHaiba 1I

I believe so! You can also just memorize that a certain shape will have particular angles.

Nada AbouHaiba 1I

As of right now we haven't covered oxidation numbers! It may be in a different class of the chem series.

Nada AbouHaiba 1I

I think this is a good tool for classifying molecular shapes

Nada AbouHaiba 1I

another way of looking at this is by looking at the types of bonds.
a molecule is most likely polar is has polar bonds, as in an atom with high EN is bonded to something with low EN.
a nonpolar molecule will compose of nonpolar bonds is two atoms with similar EN

Nada AbouHaiba 1I

Group 13 elements are an exception to the octet rule because in order for them to get an octet they need to form 5 bonds which they're unable to do.
Then any element in the third row or lower is able to have an expanded octet because they have access to an empty d orbital

Nada AbouHaiba 1I

I don't think so just because non-metals and metals are usually charged as cations/anions, so these charges would make them ionic bonds. But! there may be exceptions that I've just never heard about.

Nada AbouHaiba 1I

In order for them to have a complete octet they would need to create 5 bonds and that doesn't happen

Nada AbouHaiba 1I

The change in energy may be negative but the Energy given off can never be negative!

Nada AbouHaiba 1I

I believe Lavelle said that we would be unable to make a definitive answer if it lies in between. We would need more information/experimentation to be able to tell which type of bond it is

Nada AbouHaiba 1I

That equation is just for the kinetic energy of the ejected electron
You can't use that equation for the energy of a photon because mass is required and a photon has no measurable mass
hope this helped:)

Nada AbouHaiba 1I

There are 2 electrons (up/down spin) per magnetic quantum. So for example n=2 l=1 Ml = -1,0,1 so for each of these three there are 2 electrons. So in this case it's a total of 6 electrons. For n=4 ... ms=-1/2 It is only 1 electron because they gave you all four quantum numbers and no 2 electrons in ...

Nada AbouHaiba 1I

By just multiplying by avogadro's number, you'll get how many electrons are in those 5 moles.
You just need to go one step further and multiple that number (the amount of electron in 5 mols) by the mass of an electron, so 9.109x10^-31 kg.

Hope this helped :)

Nada AbouHaiba 1I

When the atom is excited it is gaining energy, that is why it is able to jump to higher energy levels.

Hope this helped! :)

Nada AbouHaiba 1I

I find doing the homework problems to be really helpful and then usually before the test I'll go over the homework again and redo the problem that I had initially struggled with. Also! sometime scrolling through chemistry community is really helpful just because reading how other people understand/e...

Nada AbouHaiba 1I

Hi Julia!

There are some exceptions to the general trends in the periodic table; however, we aren't expected to know them unless they're explicitly stated.

Hope this helped :)

Nada AbouHaiba 1I

when an electron is said to be in its excited state it has been "energized" and gone up to a particular shell, higher than the one it was originally at. Something that is in its ground state then it has either gone back down from being excited or there wasn't any source of energy and this ...

Nada AbouHaiba 1I

Yes, that's how they could be described! and the highest possible subshell of a particular shell would be n-1!

Hope this helped!

Nada AbouHaiba 1I

The maximum value of l is n-1.
Ml is -l to l
and Ms is either +1/2 or -1/2, that's not something we can figure out.

For example, if N=2 then..
l can be 1 or 0
Ml's -1,0,1

Hope this helped!

Nada AbouHaiba 1I

okay so I dug around a little and found a post that does a good job of explaining it, I'll link it below just in case anyone else is still confused :)
viewtopic.php?t=31402

Nada AbouHaiba 1I

Hello, I'm reviving an old post. However, I don't understand why you can figure out that n2=7 using the Balmer series. Can someone please explain? I'm also confused. I thought the balmer series started at 2, so wouldn't it be: 656.3 (n2), 486.1 (n3), 434.0 (n4), 410.2 (n5)... so wouldn't the next o...

Nada AbouHaiba 1I

I'm having trouble even finding a way to post in this forum. Can someone tell me where/how I can post a question? I agree I am having trouble with that too but in the upper left side with the three lines next to "Quick Links" shows a tab that you can view "your posts" So you go ...

Nada AbouHaiba 1I

I believe Lavelle said in class that we will never need to calculate anything using this equation, but instead we learned it in order to give us background about how it was utilized when the different orbitals were discovered.

Nada AbouHaiba 1I

This depends on whether or not one photon of light has enough energy to eject one electron. If this one photon has enough energy then by increasing the intensity of light you then increase the amount of photons, which then increases the amount of electrons that are ejected (1 photon per electron sor...

Nada AbouHaiba 1I

When you're given the amount of energy of light used to eject an electron it's important to know if there's any excess energy in the form of kinetic energy given off. If there isn't any excess kinetic energy given off then we know the energy of the incoming light is equivalent to the energy of that ...

Nada AbouHaiba 1I

Yes, you can! Depending on the color it usually has a range of wavelength it falls under - I'll attach a photo I found online that shows the different colors and their corresponding wavelengths

I hope this helps!

Nada AbouHaiba 1I

Okay this is kinda dumb, but hey whatever works, right? My (Mega) . 10^6 Kitten (Kilo) 10^3 Ben (Base) Null Doesn't (Deci) 10^-1 Count (Centi) 10^-2 Much (Milli) 10^-3 Money, (Micro) - you just have to remember here micro comes after milli 10^-6 No (Nano) 10^-9 Problem! (Pico) 10^-12 basically I thi...

Nada AbouHaiba 1I

I've also heard that we aren't expected to be able to identify different states of matter in the equation (like whether or not it's aqueous using solubility rules) so I don't think it'll be required of us!

Nada AbouHaiba 1I

So first things first you're going to want to convert to frequency using the c=wavelength*frequency equation. Then we will put these values in the Rydberg equation and for "n" you would use 1! I've also found this question worked out numerically and I'll go ahead and link that below. https...

Nada AbouHaiba 1I

I don't think we're ever going to have to identify that a competing reaction is going on. The main reason they mentioned that was to show why the actual yield is less than the theoretical yield (because the reactants can be used for something other than the main reaction going on).

Nada AbouHaiba 1I

When it comes to finding the mass percent composition of a substance how many sig figs should we round to? Or is there a certain amount of decimal places we need to round to? Or does it not make a difference? I've noticed that in the textbook solution they'll usually do up to two decimal places, but...

Nada AbouHaiba 1I

I believe you'd just write out the complete chemical equation and the do total number of moles final - total number of moles initial to find out the net number of moles.
Also, when writing out the combustion remember that the other reactant is oxygen and the products are CO2 and H20
Hope this helped!

Nada AbouHaiba 1I

Hello! Okay, to start off #23 is a dilution problem, so we know we want to get it into a M1V1=M2V2 equation. First we're going to solve for M1 andyou're going to want to convert the grams of KMnO4 into mols (so divide by the molar mass). 5.00 g KMnO4/58.04 g/mol = .0316 mol KMnO4 Then you're going t...

Nada AbouHaiba 1I

Hello, I just made my account and I cannot find where to post my questions? Please help. Hi Sarah! So you go to the main page and scroll down until you see the section labeled "Chem 14a", then go ahead and click on the subject matter that pertains to your question. Then you'll get to a pa...

Nada AbouHaiba 1I

I found the answer to this question on an earlier post and I believe they did a good job of explaining it! I'll go ahead and link it below :)

viewtopic.php?t=5565

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