CHEMISTRY COMMUNITY

Madeline Motamedi 4I

Just pay attention to sig figs in the question if the rate law is what you are speaking about.

Madeline Motamedi 4I

Yeah I actually believe Dr. Lavelle actually brought this up in class, you don't need to write the phases in the rate law however if you do it should be fine as one of the solutions for the final review had phases.

Madeline Motamedi 4I

Yeah I believe for our purpose we are usually assuming A is just 1.

Madeline Motamedi 4I

Hi I believe OH isn't an intermediate and its fine to have it in the rate law.

Madeline Motamedi 4I

Situations involving a catalysts make the rate independent of reactant concentration.

Madeline Motamedi 4I

Mathematically since you multiply k by the concentration of your reactant and the exponent is 0, this term doesn't affect the final answer.

Madeline Motamedi 4I

In these situations, it doesn't matter how much the concentration of our reactant increase, the rate will be limited by the k.

Madeline Motamedi 4I

I don't believe you can determine spontaneity based off of reaction rates.

Madeline Motamedi 4I

So I believe you can determine if a reaction is favorable thermodynamically and kinetically. I don't think we have learned it yet but I know there is a large activation energy barrier that prevents the change from diamond to graphite.

Madeline Motamedi 4I

Like the comment above, use the general equation for delta T= delta H - T times delta S. Plugging in known values into this equation will help you determine whether you have a positive or negative delta G. From there you can determine whether its spontaneous or not.

Madeline Motamedi 4I

A reaction that absorbs heat and decreases entropy results in a positive delta G which is not favorable and thus not spontaneous. If the reactants have more free energy than the products do, then you have a negative delta G. This is a favorable situation which is why it is considered spontaneous.

Madeline Motamedi 4I

Remember that although you are using delta G= delta H-T times delta S, you can still replace components of the equation with other equations you learned previously. So, you could find an equation for delta H that includes moles and any other given information. Then you could substitute this value wi...

Madeline Motamedi 4I

Like the example during class, Prof. Lavelle told us if we were to have a system being held with a lid and the pressure of the system is 2 atm while the surroundings are 1 atm if we were to release the lid, it would shoot upwards. That is an example of an irreversible system.

Madeline Motamedi 4I

I like to think of it in terms of the system. If you're the system and you're doing work, you lose energy (negative w). If your system is doing work it's losing energy. If work is being done on the system its gaining energy.

Madeline Motamedi 4I

C is juts the heat capacity for anything you are working with. On the other hand C sub v and C sub p is the heat capacity considering constant pressure (C sub p) and heat capacity considering constant volume (C sub v) for IDEAL GASSES. Remember the C sub v and the C sub p is used to calculate the he...

Madeline Motamedi 4I

The expansion for a reversible system is done infinitely slowly and less heat is released to the surroundings resulting in more work done. You can better understand this process by viewing the pressure volume graphs for the two types of systems. The area under the curves shows you the work done by t...

Madeline Motamedi 4I

Since the change n internal energy (delta U) is q plus w. In an isolated system you don't have any work being done so you can disregard w and that leaves you with q. So by studying isolated systems we can see how q changes on its own due to the reaction occurring.

Madeline Motamedi 4I

Keep in mind when using bond enthalpies you don't necessarily have to add every bond on the reactants side and subtract every bond on the products side. Instead you can just look at the lewis structures for the molecules and only add and subtract those that are going to be broken and formed.

Madeline Motamedi 4I

An extensive property is one where the quantity or amount of substance does matter. So heat capacity is an extensive property while specific heat capacity is an intensive property. The heat capacity just measures how much energy you need to raise the entire observed substance by 1 degree C. And that...

Madeline Motamedi 4I

I think the best way to approach any one of these problems is to first identify what your system is. In this case, the air inside is the system so you must look to see what is being done to it or by it. If the gas's volume expanded it pushes the piston and thus it is the one doing work. On the other...

Madeline Motamedi 4I

I'm not exactly sure which method we will be asked to do although I do know that using bond enthalpies to find delta H is not as accurate. So try to stay away from that one if looking for a more precise answer. I have a feeling we are going to be prompted to solve the problem in a certain way so jus...

Madeline Motamedi 4I

The pH tells you how acidic or basic something is. So the values range from 0-14, 0 being the most acidic and 14 being the most basic with 7 being neutral. As you move away from the number 7 either going up or down, you become more acidic or more basic. You find it by finding the concentration of hy...

Madeline Motamedi 4I

You are always adding a the change to the concentrations, whether thats positive or negative, it doesn't matter, you will find out after solving. But in terms of weather you are subtracting the x or adding the x, if you are starting with all reactants and no products, you are going to have to subtra...

Madeline Motamedi 4I

So for your reaction, you can find the formula for your Ka. Once you set up your ice table with your initial values you can grab those values and plug them into your equation for Ka (concentration of products over concentration of reactants). Since your Q value tells you the initial (or current) sta...

Madeline Motamedi 4I

I like to think of it like this: the specific chemical reaction has its own Kc it wants to maintain. So think of the whole reaction as a fraction that needs to maintain its proportions. If the top of the fraction represents concentration of products and the bottom represents concentration of reactan...

Madeline Motamedi 4I

You will notice that Kp only is calculate for reactions where the products and reactants are all gasses. If they aren't, you're most likely going to be dealing with Kc.

Madeline Motamedi 4I

Write out the reaction. Once you write it out you'll find that you end up with most likely 2 products one of which is your conjugate base or conjugate acid and the other is H30+ or OH-.

Madeline Motamedi 4I

A tip for converting this is you can actually set each of those to different variables that you assign on your own so you can see what's going on more easily. Then when you're done convert the variables back. Try doing this once so you can see the process clearly then you should be good!

Madeline Motamedi 4I

In simple terms, K represents the ratio of concentration of products over concentration of reactants. You have a fraction and if the number on top is larger, your K will be really big, versus if the number on the bottom is really big then your K is going to be really small. So if your K is a really ...

Madeline Motamedi 4I

Keep in mind that once you find K, the amounts of the products and reactants can vary although the ratios of the concentrations of reactants and products will stay the same for a certain reaction.

Madeline Motamedi 4I

For clarification, look at the second picture of the above reply. As you can tell, there is a central atom with other molecules attached and branching outwards from there. The initial bonds coming from the central atom create your coordination sphere. Everything else branching out form that initial ...

Madeline Motamedi 4I

I believe a coordination number is based on the number of regions of electron densities there are around a given central atom. So triple bonds or double bonds would count as one region of electron density and one lone pair is one region of electron density. Add the number of regions of electron dens...

Madeline Motamedi 4I

I have the sixth edition although I believe the idea behind this is that you want to place the element that the bond is going to be associated with next to the bond you drew. This is simply to signify where the bond is occurring within the atom.

Madeline Motamedi 4I

An additional way to go about this is by first drawing the lewis structure, then actually going to classify which VSEPR formula it falls under for example, AX2. From there you can memorize or start to familiarize yourself with which structures and shapes correspond to which formulas and determine th...

Madeline Motamedi 4I

So how I interpreted this is that you use your lone pairs to draw the correct shape although once you've classified the shape using the amount of regions of electron density you ignore your lone pairs. Whatever you are left with is what you will use to name the final structure.

Madeline Motamedi 4I

Nitrogen has the lowest ionization energy and the least electronegativity therefore it falls in the center of the structure.

Madeline Motamedi 4I

It is true that there are three lone pairs around the central Atom, although in this case I believe they repel equally causing the linear shape. I believe if you look at the VSEPR formula, AX2E3, simply knowing the shape for that formula, it's linear.

Madeline Motamedi 4I

Lone pairs have a large repulsion to other lone pairs which is something to keep in mind when determining the shape of a molecule.

Madeline Motamedi 4I

It's good to get in the habit of drawing the lewis structure and as you do them more and more it'll be easier and you'll become faster at drawing them.

Madeline Motamedi 4I

I believe the easier way would be to memorize the angles associated with different VSEPR Models and then explain whether in that case the angles are larger or smaller depending on the atoms involved in the molecule.

Madeline Motamedi 4I

Since the resonance structures represent a "sum" or a combination of the structures drawn, the bond length is almost like a combination of the bond lengths drawn. So even though it may seem as if there is a double bond and a single bond in another location, they can still have the same len...

Madeline Motamedi 4I

Since the exceptions would allow for the elements to fill up their d orbital it would affect the number in the valence shell.

Madeline Motamedi 4I

Sometimes they might give you the bond length and based on the resonant structures you can assume that the bond lengths are the same for all the bonds for example.

Madeline Motamedi 4I

It is difficult to model what is actually going on between bonds and resonance structures are a way to show the different states that the bonds are being "summed" up as.

Madeline Motamedi 4I

To find the formal charge, I've just been doing:
Number of valence electrons - the number of atoms directly surrounding the element (wether they are involved in a bond or just a lone pair or a radical)

Madeline Motamedi 4I

My question is, if the formal charge is 0 for all atoms in the bond, is are there resonance structures?

Madeline Motamedi 4I

I actually have a question regarding the bonds that may be extremely obvious to some. If you have an atom bonded to another with a triple or double bond, is the bond more strong than a single bond or does it simply just mean that they are sharing three or two electrons rather than one?

Madeline Motamedi 4I

If you were to write the electron configuration out you can look to see which electrons are in the highest #orbital. Those electrons with the highest n value, are the ones in your valence shell (the outermost electrons). You can then count how many there are using the periodic table or simply observ...

Madeline Motamedi 4I

When you take away an electron, the electron's charge becomes positive since you have more protons than electrons. This means that the pull on those remaining electrons is higher, it wants to keep its electrons. This make it harder to lose another. This is my understanding, but anyone please correct...

Madeline Motamedi 4I

The best way to learn the electron configurations, at least my my experience, is to look for trends in the periodic table and understand how each element and its position on the periodic table relates to its electron configuration.

Madeline Motamedi 4I

You would have to calculate the mass times the change in velocity to calculate the momentum (p) at times.

Madeline Motamedi 4I

When the nodal plane is in a cone shape or as we go higher and higher in orbital shapes, do we call these nodal areas planes or do we start to refer to them as areas?

Madeline Motamedi 4I

So, from what I'm understanding from this, the sub orbitals within one n are different distances from the nucleus. Am I correct?

Madeline Motamedi 4I

For the p orbital, we have x, y, and z since those are the three orientations for the p orbital. What happens with the orbitals after this? For the d orbitals we're dealing with 5 axis, so would we have v, w, x, y, z? or do we just use numbers to refer to these orbitals?

Madeline Motamedi 4I

So we know that a p orbital looks like an infinity sign (although, keep in mind that it isn't 2d, its 3d.) These two lobes, can be oriented in three different ways. They can be oriented along the x axis, y axis and z axis. Versus for an s orbital, which is shaped like a sphere, you only have one ori...

Madeline Motamedi 4I

I'm also wondering the same thing. I know we can calculate the kinetic energy through using a detector although are there any other physical signs we can look for?

Madeline Motamedi 4I

Can you explain each part of this equation and the relationships between each variable? I want to make sure I'm understanding it correctly.

Madeline Motamedi 4I

I'm still a bit confused as to wether light acts as a wave or not and in what situations this happens. Also, which situation would we be dealing with more often?

Madeline Motamedi 4I

I believe so far we've only really conveyed from kilograms to grams then from grams to milligrams as well as liters to milliliters and also moles to mmoles, am I correct?

Madeline Motamedi 4I

This is a bit embarrassing, but I was wondering if anyone knows of the the best way to memorize conversions between various units. Do you guys have any videos or resources for me to memorize?

Madeline Motamedi 4I

I also had this same concern when doing the post survey for the video modules. There was a situation where I was very close to a whole number so I created my empirical formula although it was wrong. I'm hoping I will learn how to create the ratios with more practice.

Madeline Motamedi 4I

So there are four scenarios 1. Low accuracy and low precision 2. Low accuracy and high precision 3. High accuracy and low precision 4. High accuracy and high precision Obviously we are shooting to get high accuracy and high precision; although, in the event that your data isn't, would it be worse to...

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