CHEMISTRY COMMUNITY

Chinmayi Mutyala 3H

You look at the overall reaction. Anything that isn't in the overall reaction is an intermediate.

Chinmayi Mutyala 3H

The first step was a fast step and right after it was the slow step. Since the slow step takes longer to happen, in that time, for the fast step, there is a buildup of products which will go in the reverse and form reactants. This is the assumption we're making in the pre-equilibrium approach. In th...

Chinmayi Mutyala 3H

Gibbs Free Energy relates to thermodynamics because it represents the amount of energy that can be used for the reaction. The equation for Gibbs Free Energy relates it to enthalpy, temperature, and entropy. It doesn't have to do with kinetics because it doesn't tell you how fast or slow the reaction...

Chinmayi Mutyala 3H

I think it would probably be specified in the question and they should be in the same phase.

Chinmayi Mutyala 3H

I feel that. It's so hard to stay motivated right now but I've just been taking it day by day. Good luck on finals everyone!!

Chinmayi Mutyala 3H

We aren't going to be plotting graphs since the test is multiple choice but he probably wants us to know what's on the y-axis for the zero, first, and second order graphs.

Chinmayi Mutyala 3H

Every reaction needs to reach a transitional state which requires a large amount of energy and that's called the energy barrier or activation energy. When we talk about a reaction being spontaneous, we're only talking about the difference in free energy between the products and reactants so we compl...

Chinmayi Mutyala 3H

The Integrated Rate Law lets you figure out the rate of the reaction at a single time. In other words, it gives you the instantaneous rate. The reason it specifies the products are being removed is because if the products were part of the reaction, the reverse reaction could occur and the reactant c...

Chinmayi Mutyala 3H

Yes, that's correct! Electrons move from left to right.

Chinmayi Mutyala 3H

Oxidizing agent is the one getting reduced and you determine strength by looking at standard reduction potential so if it is low, it's a poor oxidizing agent.

Chinmayi Mutyala 3H

delta U is 0 for an isothermal reversible expansion since the amount of energy lost from work is replaced by heat. q= -w

Chinmayi Mutyala 3H

When work is done on the system, it is positive and when work is done BY the system it is negative since the system is using energy to do work. Hope that makes sense!

Chinmayi Mutyala 3H

H and q are the same under constant pressure. Otherwise, you'd get to different values for each. I'm not entirely sure how they're different conceptually.

Chinmayi Mutyala 3H

it depends on whether pressure is constant or not. You can use -P(delta)V when external pressure is constant and the other when it isn't.

Chinmayi Mutyala 3H

This depends on whether pressure is constant or not. You can use -P(delta)V when external pressure is constant.

Chinmayi Mutyala 3H

Enthalpy is equal to heat under constant pressure.

Chinmayi Mutyala 3H

deltaH equals q under constant pressure.

Chinmayi Mutyala 3H

I'm pretty sure they're the same thing.

Chinmayi Mutyala 3H

Internal energy (U) is a state property because you'll get the same result no matter what path you take. It only matters what your initial state was and what your final state was.

Chinmayi Mutyala 3H

An extensive property is a property that changes with the size of the sample. Heat capacity is an extensive property because if you have more of the sample, it will take more heat to change the temperature. An intensive property is a property that doesn't change with the size of the sample. Specific...

Chinmayi Mutyala 3H

The work done is replaced by heat so Q = -W. Since delta U = Q + W, it would be 0.

Chinmayi Mutyala 3H

In his examples in class, he just uses 273 so I think that should be fine.

Chinmayi Mutyala 3H

When there's pressure and volume involved in the calculation, I use the R constant that includes atm and L in the units. You just have to check the units and make sure it cancels out to get you the final units you need.

Chinmayi Mutyala 3H

There are no units. It's just a measure of the number of possible states I believe.

Chinmayi Mutyala 3H

If delta G is negative, the forward reaction is favored so it will form products.

Chinmayi Mutyala 3H

2x is considered to be the concentration of that compound. When you calculate K, you do each concentration to the power of it's stoichiometric coefficient so it would be (2x)^n. The exponent would be on the 2 as well.

Chinmayi Mutyala 3H

Percent protonation/deprotonation compares the H3O+ concentration or OH- concentration with the initial concentration. Percent ionization compares the ion concentration to the initial concentration. They can be similar. It depends on the context.

Chinmayi Mutyala 3H

I'm taking Chem BL right now. You can honestly take it whenever, not just with Chem 14B. There are concepts from Chem 14A that you need to remember but nothing too complex. It'll be pretty easy to review those so you don't need to take it right after.

Chinmayi Mutyala 3H

Al has a charge of 3+ and is able to form a coordination compound. When it binds to water, it weakens the existed OH bond in water which leads to an H+ being released.

Chinmayi Mutyala 3H

It would shift to the left since by compressing the system, you are increasing the pressure. The left has less moles of gas. By shifting the equilibrium to the left, more solid reactant would be made and decrease the pressure of the system.

Chinmayi Mutyala 3H

Partial pressure refers to how much of the total pressure is occupied by one gas so we call it partial pressure of that gas since that is occupying that "part" of the pressure.

Chinmayi Mutyala 3H

When K is less 10^-4 is what Lavelle said in lecture but he also said to check the percent ionization so divide x by the initial concentration and if it's less than 5%, the assumption is okay.

Chinmayi Mutyala 3H

Simplifying it probably refers to using the assumption.

Chinmayi Mutyala 3H

This is also a strong acid.

Chinmayi Mutyala 3H

No, this information will probably be given in the question on the test since it's not on our constants & equations sheet.

Chinmayi Mutyala 3H

Less than 10^-3 but Lavelle said when k is less than 10^-4 to be safe; also if you divide x by the initial concentration and get less than 5%, you know that your assumption is ok.

Chinmayi Mutyala 3H

Yes, endothermic will be positive since it is absorbing heat and exothermic will be negative since it is releasing heat.

Chinmayi Mutyala 3H

Yes, equilibrium composition means equilibrium concentrations. You use the ICE box when you are given initial concentrations and you need to find K or the equilibrium concentrations.

Chinmayi Mutyala 3H

You would have the check the pressure units to figure out what constant to use and you don't need to memorize them because they can all be found on the Constants & Equations Sheet on Lavelle's website.

Chinmayi Mutyala 3H

It identifies it as a weak base so you would need to find Kb. You can do this by using the formula Ka x Kb = Kw so you would just divide 1.0 x 10^-14 by the given Ka.

Chinmayi Mutyala 3H

If you take the -log of the hydronium ion concentration (H3O+, or H+), you will get pH and if you take the -log of the hydroxide ion concentration (OH-), you will get pOH.

Chinmayi Mutyala 3H

If the pH is less, the solution the acid is in is more acidic so it won't deprotonate in the solution. The solution already has a high hydrogen ion concentration so it won't be able to add hydrogen ions easily which is why it remains neutral.

Chinmayi Mutyala 3H

You only look at molecules in the gas phase since those are the one's whose concentration is affected by a change in pressure. However, this change in pressure must cause a change in the volume of the container. If the pressure is changed by adding another gas, it won't affect the concentrations of ...

Chinmayi Mutyala 3H

Yes! This is because when you multiple the two, the other reactants and products will cancel out and you will be left with [H3O+][OH-] which is the equation for Kw.

Chinmayi Mutyala 3H

If Q = K, the reaction is at equilibrium.

Chinmayi Mutyala 3H

Amphiprotic means it can act as both a proton donor or acceptor. Water is an example because it can donate an H+ and become OH- or it can accept an H+ and become H3O+.

Chinmayi Mutyala 3H

When the temperature changes, K will change. These problems usually give us a temperature to let us know that the reaction is occurring at the same temperature as K is being reported at.

Chinmayi Mutyala 3H

You look at the units of R. It depends on what units of pressure you're using.

Chinmayi Mutyala 3H

For me, it's easy to understand the relationship between the two when you look at a balloon. If you have a large balloon (high volume), the particles inside have a lot more area to move around in which is why there's lower pressure. If the balloon is smaller, the particles are moving around in a sma...

Chinmayi Mutyala 3H

P: pressure
V: volume
n: moles
R: R constant
T: temperature

Chinmayi Mutyala 3H

K is products over reactants so if you use the reverse reaction, K would be the reactants over the products. That's why K (reverse) is 1/K (forward)

Chinmayi Mutyala 3H

Partial pressures is the pressure of each individual gas. When all the partial pressures are added up, you get the total pressure. It can be convereted to concentration (which is measured in mol/L) using the ideal gas law: PV = nRT --> you solve to get n/V which would be mol/L.

Chinmayi Mutyala 3H

Whether it is tetrahedral or octohedral depends on the number of ligands it is bonded two, 4 for tetrahedral/square planer and 6 for octohedral.

Chinmayi Mutyala 3H

This is what I got from the lecture. If the lone pairs are on the top and bottom, they'll be 90 degrees away from all the other bonds which is not the best for them to be as far away as possible. You want them to be as far away are possible so the lone pairs are kind angled in the axis in three dime...

Chinmayi Mutyala 3H

2F has to do with the Molecular Shape unit so it wasn't on Midterm 2. Midterm 2 had upto 2D which had to do with Chemical Bonds.

Chinmayi Mutyala 3H

Thank you so much Kate! I learned a lot from you this quarter!!

Chinmayi Mutyala 3H

Thank you so much Matthew! I learned a lot from you this quarter!!

Chinmayi Mutyala 3H

During the workshop I attended, the UA said to know about cisplatin and the heme complex.

Chinmayi Mutyala 3H

They don't contradict each other because s orbitals have lower energy, so that's why electrons are added to 4s before 3d and since they have lower energy, they are easy to remove so when a cation is formed, they are removed from 4s orbitals.

Chinmayi Mutyala 3H

Based on the outline, I don't think we'll be asked this on the test.

Chinmayi Mutyala 3H

The hybrid orbital specifies the principal quantum number (n) of the orbital.

Chinmayi Mutyala 3H

Whether it's a double bond or a single bond, there is only one hybrid orbital that will participate in a sigma bond. The other bonds come from overlapping p-orbitals (pi bonds). This is why you only count it as one region of electron density.

Chinmayi Mutyala 3H

The coordination number is the number of bonds the transition metal forms.

Chinmayi Mutyala 3H

Hybridization is the mixing of orbitals and it happens when an atom binds with other atoms so you can't tell what the hydridization will be from the electron configuration of the atom alone.

Chinmayi Mutyala 3H

A delocalized pi bond is present in a molecule that has resonance because the e- are delocalized.

Chinmayi Mutyala 3H

Single bonds are sigma bonds and double bonds have one sigma and one pi bond. The difference is that in sigma bonds, the ends of the orbitals overlap and in pi bonds, parallel orbitals overlap side to side.

Chinmayi Mutyala 3H

No, it would need to have a d-orbital and atoms in 3s don't use the d-orbital.

Chinmayi Mutyala 3H

You don't need to know the values, you should just know the periodic trends.

Chinmayi Mutyala 3H

It doesn't need to be bent. You have to check whether the dipole moments cancel out on the sides of the molecule.

Chinmayi Mutyala 3H

No, hydrogen bonds require hydrogen bonding with an electronegative atom.

Chinmayi Mutyala 3H

The more electronegative atom would have the negative formal charge.

Chinmayi Mutyala 3H

Yes they are. Intermolecular forces hold separate molecules together and intramolecular forces are forces that hold a single molecule together.

Chinmayi Mutyala 3H

Electronegativity increases as you go to the right and it decreases as you go down on the periodic table.

Chinmayi Mutyala 3H

I don't think we need to know about the specifics of how we got the value exactly; just that + means upward spin and - means downward spin.

Chinmayi Mutyala 3H

He was referring to the octet rule. It has 5 valence electrons but it wants 8.

Chinmayi Mutyala 3H

They happen for all: both nonpolar and polar molecules.

Chinmayi Mutyala 3H

There is not that great of an electronegativity different between C and H for it to be a polar bond.

Chinmayi Mutyala 3H

I'm not sure what your question is asking exactly but just a trend to know is that double bonds are shorter than single bonds.

Chinmayi Mutyala 3H

I think subshell is the combination of the shell and the which orbital it is (which would be represented by l). I'm not sure if this is the best way of explaining it.

Chinmayi Mutyala 3H

It probably depends on the situation but I would say it's probably more stable to have it spread out since you don't want a lot of electrons in one region since that leads to electron-electron repulsions.

Chinmayi Mutyala 3H

There are certain times when it can be violated such as P and I which are exceptions because they have an expanded valence orbital. These are just exceptions we'll have to memorize.

Chinmayi Mutyala 3H

Pi bonds do not use hybridization. They are formed by overlapping p-orbitals.

Chinmayi Mutyala 3H

The 2p orbital is on the second shell which means the radius is increasing. Since there are more electrons, there is more repulsion between them so the radius increases.

Chinmayi Mutyala 3H

Hi! This question has to do with Heisenberg Uncertainty/Indeterminacy principle.

Chinmayi Mutyala 3H

I feel like that should be true since higher frequency means more energy but maybe it's because the frequency isn't enough to eject the electron. I'm not sure if there were any additional details given in the question.

Chinmayi Mutyala 3H

This works for hydrogen only. There are some "corrections" that need to be made when doing this for other elements. I'm pretty sure we're just focusing on hydrogen but there's some information on this in the textbook.

Chinmayi Mutyala 3H

This would be determined based on what information is given in the question.

Chinmayi Mutyala 3H

Any post or reply counts for participation points. You can click on your name to see how many posts you did and the day they were posted.

Chinmayi Mutyala 3H

Hi! We will not be tested on the particle in a box on the midterm. Dr Lavelle said in his lecture today that our midterm will only go up to the last slide he did which introduced the different atomic orbitals and did not discuss a particle in a box. In general, our midterm will only contain all the...

Chinmayi Mutyala 3H

Samin Kabir wrote:Hello,
When we have to balance chem equations do we have to include the states of matter in all components. Additionally how do we know what state of matter a reactant/product is in?

I would add it in just to be safe and it should be given in the question.

Chinmayi Mutyala 3H

Hi! What makes something a solvent or a solute? How can we just know? I know that solutes are dissolved in solvents to create a solution such as salt and water, respectively. However, if given two items, how would I know? What makes something soluble I guess? I don't think his questions will go sup...

Chinmayi Mutyala 3H

905579227 wrote:Do you guys know what will be allowed on the midterm? Also do we have to memorize the conversions or will that be provided.

The conversions will be provided. I would say you should memorize the electromagnetic spectrum and the prefixes like giga- , mega- etc.

Chinmayi Mutyala 3H

I believe we just need to know the relationship between Schrodinger's and orbitals.

Chinmayi Mutyala 3H

I'm not entirely sure but since they've given up all their energy to the electron, I think they just get destroyed in the process.

Chinmayi Mutyala 3H

hi! Mine is kind of weird. I do: the G reat M aster K night D id C ommit M urder U nder N o P retenses. G - giga M - mega K - kilo D - deci C - centi M - milli u - micro n - nano p -pico It is so weird that I have no trouble remembering it. I don't remember where it is from, but it works. :) Whoa! ...

Chinmayi Mutyala 3H

Your final answer should have whole number coefficients.

Chinmayi Mutyala 3H

I believe this depends on how strict your TA is on sig figs but I would round to the correct number of sig figs on your final answer based on the numbers given to you in the problem just to be safe.

Chinmayi Mutyala 3H

Protons and neutrons have similar masses and electrons have a much lower mass than protons and neutrons.

Chinmayi Mutyala 3H

Also, for the external camera, I believe it's okay to use a phone.

Chinmayi Mutyala 3H

I'm not sure if this is what you're asking but once you find the mass of carbon and hydrogen, you can subtract that from the mass of the original compound to find the mass of oxygen.

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