CHEMISTRY COMMUNITY

Malakai Espinosa 3E

Also a difference to note, catalysts can be included in your rate law if they are in the slow step, but intermediates cannot!

Malakai Espinosa 3E

Did you check to make sure your units lined up? If you didn't covert KJ to J that could be an issue.

Malakai Espinosa 3E

This is applying the pre-equilibrium concept!!! We don't want to have intermediates in our rate law when dealing with a fast step preceding a slow step, so it's in our best interest to isolate the intermediate in the first step and substitute the concentrations. I believe Lavelle talked about this i...

Malakai Espinosa 3E

Jeffrey Hablewitz 2I wrote:Is this formula given on the formula sheet? Or are we just expected to memorize it?

The equation deltaS=q/t is given on the equation sheet, so it's basically applying this equation!

Malakai Espinosa 3E

Just like Stuti said, we know it's endothermic because energy/heat is required in order to break bonds. Also for D, the catalyst does not shift the equilibrium constant in any way. The catalyst's function in kinetics vs thermodynamics is not really too related!

Malakai Espinosa 3E

In a salt bridge, the anions from the salt move towards the anode and the cations from the salt move towards the cathode. This occurs because we need to stabilize the charges of each half cell.

Malakai Espinosa 3E

I also believe if you look through the appendixes there are half reactions with values for both of them.

Malakai Espinosa 3E

We can look at experimental data or a graph to determine the order of reactions.

Malakai Espinosa 3E

With integrated rate laws, we only want to analyze the reaction in the forward direction. Thus, there needs to be some way in which the product is removed so our reaction isn't affected by any reverse reaction.

Malakai Espinosa 3E

I feel like on the test it may give you the graphs and ask to analyze them, especially since we can't make the graphs on an online multiple choice.

Malakai Espinosa 3E

For the rate laws we are looking at, we are only considering the rate of reaction in the for the forward direction, hence why we look at the initial reaction concentrations.

Malakai Espinosa 3E

You get pseudo K for a specific reactant when making the other reactants in large excess. The relationship between k and k' is k=k'/([B]0 [C]0)

Malakai Espinosa 3E

You need a solid metal to be a conductor of electrons. A galvanic cell cannot take place if there are not conductors on both sides of the reaction

Malakai Espinosa 3E

Anions will flow towards the anode because that half-cell is slowly becoming more positive, so the anions balance the charge out. This is the same concept for the cathodes, but cations are neutralizing the more negative charge.

Malakai Espinosa 3E

You want to look at the oxidation numbers of Cu and Cl in CuCl, and this will tell you which half reactions you should use.

Malakai Espinosa 3E

The higher the standard reduction potential, the more likely a reaction is to take place in it's reduced direction.

Malakai Espinosa 3E

Q is a more broad variable, as it can represent the ratio of P and R at any given time in the reaction, whereas K is only at equilibrium.

Malakai Espinosa 3E

Open system can exchange matter and energy, closed system can only exchange energy, and isolated system cannot exchange matter nor energy.

Malakai Espinosa 3E

If delta H and S are both positive, the reaction will be spontaneous at high temperatures.

Malakai Espinosa 3E

The anode is where oxidation comes up, and this reduces the cathode.

Malakai Espinosa 3E

I believe there has to be a battery/galvanic cell for chemical reactions to create electrical energy.

Malakai Espinosa 3E

If work is done by the system, it's negative. If work is done on the system, it's positive.

Malakai Espinosa 3E

temperature will pretty much always be in kelvin, as the R constant will be in Kelvin or the entropy will be in J/K.

Malakai Espinosa 3E

This helps us solve the equation. Another way I like to solve this problem is to do set an inequality where deltaS(T)>deltaH, and then solve for T.

Malakai Espinosa 3E

The fundamental particles are what you described: protons, neutrons, and electrons.

Malakai Espinosa 3E

Like the others have said delta G naught is at standard conditions 298k and 1 atm. In general whenever there is a naught next to a variable, this signifies that the system is at standard conditions.

Malakai Espinosa 3E

If you watch today's lecture he gives us the equation delta G naught = -RTlnK, which applies when the reaction is at equilibrium since delta G = 0

Malakai Espinosa 3E

The units for K cancel so there is actually no units. It merely tells us the ratio of products and reactants relative to one another.

Malakai Espinosa 3E

In the question it's asking for the change in energy, and that the temperature of the gas was raised. Energy(specifically heat) is being put into the system(the gas) so the internal energy is being raised. The problem also states that it's at constant volume, so the work would be zero.

Malakai Espinosa 3E

I endorse UA sessions so much they're really helpful. I like the Step-Up sessions(especially Hannah's) the most because they're smaller and you can get your questions explained in a more in depth manner.

Malakai Espinosa 3E

Okay like the textbook definition of entropy of disorder, but I like to think of it more as the amount of chaos like in terms of molecules moving around or the amount of molecules. It just help sme understand better in my head.

Malakai Espinosa 3E

Like the others said, I believe U is extensive because it depends on how many molecules there are.

Malakai Espinosa 3E

for a), lactid acid provides more H3O+, so use that to solve. For b) Hb is increasing in the reverse reaction

Malakai Espinosa 3E

The way I approached it was by first multiplying the top reaction by 2, so that there are 4NO2, which will cancel out with the 4NO2 in the bottom reaction. I then created the overall reaction and added their delta H(keep in mind the top delta H is multiplied by 2). After that I used the Hf of NO (an...

Malakai Espinosa 3E

if Ka or Kb is less than 1E-4 then we can use the approximation, since the value of x would be really small.

Malakai Espinosa 3E

I think understanding the heating curve will be a good portion of what we need to know, but it wouldn't hurt to know the steam concept because he went over lecture.

Malakai Espinosa 3E

The way I approached it was by first multiplying the top reaction by 2, so that there are 4NO2, which will cancel out with the 4NO2 in the bottom reaction. I then created the overall reaction and added their delta H(keep in mind the top delta H is multiplied by 2). After that I used the Hf of NO (an...

Malakai Espinosa 3E

Like the others said I think he meant to separate his sentence and was referring to I2 as Iodine being a liquid.

Malakai Espinosa 3E

I believe based on what information you are given will determine the best method to solve the problem!

Malakai Espinosa 3E

The pressure is increasing but the volume isn't decreasing, so the concentrations remain the same since the inert gases are very unreactive and won't change.

Malakai Espinosa 3E

You are right! The temperature stays constant during phase changes because the heat energy is going to the formation and breaking of bonds.

Malakai Espinosa 3E

I use the acronym Mr. HOFBrINCl (like hoff-brinkle) to memorize the diatomic molecules it's been pretty helpful for me.

Malakai Espinosa 3E

Because the two concentrations are the same, we can both define the values as x. Thus, Kw would be x^2. In order to find the concentration of either, we need to square root the Kw that is given.

Malakai Espinosa 3E

In the second equation H2O is acting as the base of the reaction, whereas in the first equation H2O is merely breaking the Ba(OH)2 apart.

Malakai Espinosa 3E

Pressure only applies to molecules in the gas phase.

Malakai Espinosa 3E

The formula you listed can be used to find the pH/pOH for all acid base reactions, not just strong ones. Weaker acids and bases will have higher values.

Malakai Espinosa 3E

pH and pOH are from the concentrations in water I believe because an acid/base has to be put in water in order to form OH or H3O.

Malakai Espinosa 3E

The rates of the forward and reverse reactions are equal at equilibrium!

Malakai Espinosa 3E

Usually the given concentration will change with a -x because the reaction will proceed in the opposite direction in order to reach equilibrium.

Malakai Espinosa 3E

If H2O is in the liquid or solid phase, then we don't include it in the ICE table. However, if H2O is in the gaseous phase, then we do include it in the table.

Malakai Espinosa 3E

Like everyone said before, if we increase the volume of the reaction then the overall concentrations will decrease on both sides. As a result of Le Chateliers principle, the reaction will then stabilize by favoring the direction where there are more mols of gas.

Malakai Espinosa 3E

The change in pressure only increases the concentration if the change is a result in a decrease of volume. concentration=moles/volume. The moles stays constant but the volume decreases, so the concentration increases.

Malakai Espinosa 3E

Ka is a value for the ratio of products to reactants. pKa helps us tell the strength of a weak acid, and this is determined by taking the negative log of the Ka value.

Malakai Espinosa 3E

Like the others have said, I don't believe there is a specific conjugate acid or base for this reaction since neutralizations between a strong acid and strong base create a salt plus water.

Malakai Espinosa 3E

In terms of coordination compounds, we wont learn how to tell the difference between square planar and tetrahedral until other quarters of chem, but for VSEPR the difference is that square planar has two extra sets of lone pairs in addition to the 4 bonded atoms.

Malakai Espinosa 3E

Some of the elements that don't follow the octet rule are B, Be, and Al.

Malakai Espinosa 3E

HClO is more acidic because in the anion, Chlorine is more electronegative which stabilizes the anion more.

Malakai Espinosa 3E

Monodentates are ligands that can only bind to the TM at one point, whereas polydentates are ligands that can bond to the center TM at multiple locations.

Malakai Espinosa 3E

Electronegativity is helpful in determining the relative acidity of molecules. A molecule is more acidic when there are higher electronegativty atoms because they delocalize electrons and stabilize the negative charge.

Malakai Espinosa 3E

Like Gerardo said, pH involves the negative log of the concentration of H+ ions in a solution whereas pOH is for the negative log of the concentration of OH- ions. In a way they are complementary to one another but pH is what we mainly use.

Malakai Espinosa 3E

OH2 I think is just helpful to write sometimes that way we know the molecule is binding to the Oxygen atom in water, and not one of the hydrogen atoms.

Malakai Espinosa 3E

It's not necessarily that the HBr molecule is the lewis acid itself, but it is providing the Lewis acid which is the H+.

Malakai Espinosa 3E

Lone pairs will be placed to maximize their distance from other lone pairs and bonds. It helps to have a physical model like Dr. Lavelle showed us in class, as if the bonds are 3D you can clearly tell where the lone pairs should be placed.

Malakai Espinosa 3E

In my opinion, I think visualising the shape of a molecule best helps me understand polarity. It's like a game of tug of war. If both sides are equal in strength, the rope won't move(nonpolar), however, if one side is stronger than the other, than the rope will gravitate towards that side(polar). Al...

Malakai Espinosa 3E

When a molecule is bent with 1 lone pair on the central atom, the electron arrangement is trigonal planar. Trigonal planar has 3 bond angles roughly 120 degrees each. That is why it would be less than 120, and not 109.5, which are the approx. bond angles of an atom in a tetrahedral arrangement.

Malakai Espinosa 3E

To find the hybridization of a given atom, simply look at the regions of electron density around the atom! This includes bond pairs and lone pairs. This will allow you to figure out the hybridization.

Malakai Espinosa 3E

If you could memorize what shapes create nonpolar molecules, that is perfectly fine. However, for more complex molecules it is also helpful to know how to draw proper lewis structures to analyze whether or not certain polar bonds cancel each other out.

Malakai Espinosa 3E

The dipoles have to cancel in a linear manner, kind of like tug of war where both sides are equal in strength. If you tried to do tug of war when both sides are not completely opposite from one another, it wouldn't really work. Hope this analogy helps :)

Malakai Espinosa 3E

I believe for sigma bond, their axises are positioned end to end, allowing for rotation. For pi bonds, the axises are side by side, meaning that they cannot rotate.

Malakai Espinosa 3E

Like everyone said, this would be a linear shape!

Malakai Espinosa 3E

I'm not sure how essential it would be to memorize the angles, but I feel like there could be questions dealing with the bond angles on the final? I think it's still helpful to know the bond angles, maybe not exactly the number but their length relative to the other shapes.

Malakai Espinosa 3E

Electron repulsion comes when negatively charged electrons cause repulsion between each other.

Malakai Espinosa 3E

Delta negative just represents the unequal sharing of the electrons in a covalent bond. Delta negative indicates the electrons are being pulled more towards that atom, and thus delta positive shows that the electrons are being pulled more away from the atom(in the bond of course).

Malakai Espinosa 3E

Dissociation energy is the energy required to break a bond. I think these values are usually found through experimentation.

Malakai Espinosa 3E

He just went over this in todays lecture! Single bonds are the longest, then double, then triple bonds. Also, the bigger the radius of an atom is the longer the bonds will be.

Malakai Espinosa 3E

I believe ionic bonds have basic characteristics of covalent bonds, but the difference in electronegativity is so large that the unequal sharing gives them also gives them ionic character.

Malakai Espinosa 3E

Can someone explain what this equation (u=|q|d) means? Why is there an absolute value sign? What does the q stand for? The equation is the equation to show the strength of a dipole. dipole=charge x distance. The q stands for the charge(delta + or -). The absolute value is there to ensure that the d...

Malakai Espinosa 3E

Having that one electron unfilled in the 4d state makes the atom less stable, so the final electron configuration actually moves the electron from the higher energy p subshell to the d subshell, that way the d subshell can be completely filled. I believe this exception applies with all the transitio...

Malakai Espinosa 3E

I think I also had this question because he never specified how we write F in electron configurations during lecture.

Malakai Espinosa 3E

Like everyone else I said I'm sure it's not a huge deal which way you write. However, I would just write it with the previous noble gas just to be safe.

Malakai Espinosa 3E

The charge outside of the square bracket demonstrates the charge of the entire molecule. Because many of the structures we model are ions, they must have a charge. Using the sign outside the bracket is how we show that charge.

Malakai Espinosa 3E

Lewis structures are a simple way for us to model what an electron structure may be. Since it is relatively simple, certain characteristics of the atom may not be seen. Like everyone said previously, resonance helps us account for something that a Lewis Structure doesn't demonstrate at first glance....

Malakai Espinosa 3E

Atomic radius decreases as you got from left to right across the same period/row. This is because there are more electrons, increasing the attraction towards the nucleus. As we go down a column, the radius increases because the electrons are being added to another energy level.

Malakai Espinosa 3E

since we have momentum in the uncertainty equation, we must use meters. This is because momentum uses velocity with SI units m/s. It is always just easier to convert nm to m.

Malakai Espinosa 3E

Yes Rydbergs equation only works for hydrogen atoms. I believe in order to use it for other elements we would have to modify the equation.

Malakai Espinosa 3E

P is simply the variable we use to represent momentum! It can be broken down as mass*velocity.

Malakai Espinosa 3E

n is the shell
l is the angular momentum(shape) of the shell
ml is the orientation of the shell, based on the axis of the nodes
ms is the spin of the electron represented as +/- 1/2

Malakai Espinosa 3E

Yes the Photoelectric experiment must be conducted under a vacuum in order for scientists to properly calculate everything. If this experiment were not conducted under a vacuum then miscellaneous particles in the air(oxygen atoms, nitrogen, etc.) would affect the ability to detect values such as the...

Malakai Espinosa 3E

The number you listed is an estimated value for the speed of light! Some people round to three while others use more specific values like 2.98. I believe Professor Lavelle is okay with us rounding to 3 though.

Malakai Espinosa 3E

Photons do have a mass! They're a little heavier than electrons. I didn't catch the part about photons not having mass in lecture, but photons, neutrons, and electrons all have mass. Hope that clarifies things. Wait my bad I thought you were talking about protons haha. Like the others said, photons...

Malakai Espinosa 3E

Electrons can only have a certain amount of energy to be within a particular energy level. We can excite electrons to higher energy levels when they absorb energy, but I believe that makes the atom unstable, so they need to go back down to their normal energy level, thus the need to release the ener...

Malakai Espinosa 3E

Like all the previous posts said, the equation would be E=hv. This equation does not directly describe the photoelectric effect, but is merely a component in explaining the experiment. E=hv actually describes the energy of a photon, which we can then use along with the work function to show the phot...

Malakai Espinosa 3E

Photons do have a mass! They're a little heavier than electrons. I didn't catch the part about photons not having mass in lecture, but photons, neutrons, and electrons all have mass. Hope that clarifies things.

Wait ignore this I misread the question.

Malakai Espinosa 3E

I think Nika did a great job of explaining it! The photoelectric experiment helps explain particle theory of light because it shows that the energy required to eject an electron is determined by a particle of a photon. There is a 1:1 ratio of photons to electrons ejected. Increasing the intensity/fr...

Malakai Espinosa 3E

Φ is described as the work function/threshold energy, but I don't believe there is an equation that we need to know in order to get the value on our own. This number is more determined through experimental data, and therefore beyond the scope of this class. If the work function is needed then the pr...

Malakai Espinosa 3E

Not only does this concept apply to light waves, but waves in general! You could apply this to sound for example. If you increase the amplitude of the sound waves, the intensity/volume of the sound you hear will be perceived as louder.

Malakai Espinosa 3E

Hey Michael,
I believe in Dr. Lavelle's lecture he mentioned how if there is not enough energy to eject electrons from the metal, the light will simply pass right through the metal. The light only gets absorbed if the energy is equivalent or higher than the threshold energy to eject electrons.

Malakai Espinosa 3E

A little acronym my AP Chem teacher told me that helps me remember is HOFBrINCl but it's pronounced like Hoffbrinkle. It's a funny name so it's easy to remember.

Malakai Espinosa 3E

So based on the responses, I gather the understanding that in order to find the amount of excess, we need to subtract the mols of product formed by the limiting reagent from the mols of product formed by the excess? Then we convert that number into grams of the excess reactant? Can someone clarify t...

Malakai Espinosa 3E

The first thing I did was figure out the molecular formulas or 2butone and 3 methyl. Every like "point" or "corner" is a carbon molecule. Every carbon molecule forms 4 bonds. So we have to imagine there are hydrogen molecules attached to each carbon to fulfill these 4 bonds. So f...

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