CHEMISTRY COMMUNITY

maldonadojs

it is not usually negative, but in some instances, we have allowed it to be negative. i think when you are not trying to make the most spontaneous and just solving for E, you should follow cathode-anode and hope it does us justice. I was also confused.

maldonadojs

this equation is the integrated rate law for first order reactions and to add to what the previous post said, you are able to graph this as just a downward straight line, because of the -k (slope) and the y intercept, ln [A]0

maldonadojs

that clears things up. thank you so much! (:

maldonadojs

I understand that a salt bridge prevents charge build up and allows for the salt ions to move in order for the reaction to continue, but does the porous disk allow charge build up? I know the solid expands in size throughout the experiment, but how does the porous disk motivate this?

maldonadojs

All reactions use activation energy, but exothermic reactions just produce more energy than they consumed. So to answer your question, having activation energy does not declare a reaction being endothermic vs exothermic. They could have the same Ea, but they depend on the initial and final energy us...

maldonadojs

It is difficult to understand integrals without the knowledge of derivatives, which is just another word for slope. I have attached two links to explain the two.
https://youtu.be/IHaK3XzTse4
https://youtu.be/rAof9Ld5sOg

maldonadojs

I am sure the question will tell us to use the k as a rate constant in kinetics in the question, whereas the k Boltzmann constant will be used to calculate the entropy. K will be given in an acids and base question. It is based off of context.

maldonadojs

With regards to the integrated rate law, this is due to the fact that when you are integrating the derivative, because of the integrated version being 1/[A] there happens to be a -1, but they divided the negative from the k over to cancel these out.

maldonadojs

Coming from the left, it is solid|g,L|aq.||aq.|L,g|solid

maldonadojs

E naught is under standard conditions. 25° C, 1 atm.

maldonadojs

The unique rate just takes coefficients into account and you multiply them to the concentration (the ones you use in an instantaneous rate).

maldonadojs

Whenever there’s no elements w the (s) subscript. That’s the element you add it to because Platinum is a non reacting element in these reactions.

maldonadojs

When you add the electrons to the right side, I believe it will be reduction and electrons to the left of the arrow is oxidation.

maldonadojs

There is importance in this because based off of the solution being acidic vs basic, you must add different components. Basic is very similar to acidic, but you would then add OH- ions to balance the half equations, whereas the acidic requires just H+ and electrons

maldonadojs

deltaU=0 when there is a constant temperature, or when temperature is 0. This is because of the equation U=3/2nRT. If constant temperature, Tf-Ti=0. However, if temperature=o, the equation would equal zero instantly.

maldonadojs

This is tied back to the equation PV=nRT. Due to 'P' and 'V' being multiplied together, they are inversely proportional. When volume goes up, pressure goes down and vice versa.

maldonadojs

The one using V2/V1 is when volume is changing and you have a constant temperature. You would use the one with the T2/T1 when temperature is changing. In this case, you would use Cv or Cp as the constant values.

maldonadojs

Delta U is 0 when you are dealing w a isothermal/reversible reaction.

maldonadojs

U(tot)=(3/2)nRT is the calculation of total internal energy. You use this equation for monoatomic ideal gases, which means just one atom. These problems will typically use the information from one of the ideal gases (aka noble gases). The questions also are typically reversible. The idea behind this...

maldonadojs

When K=1 the concentrations are the same of the products and the reactants. I am unsure as to what you mean by "intermediate," so I am sorry and I hope I answered your question.

maldonadojs

From my knowledge, reversible reactions happen simultaneously, whereas an irreversible goes in steps. For this reason, say we were to calculate the area under the curve of the two reactions, the work is greater in a reversible reaction. The graph of a reversible is a curve (which has more area under...

maldonadojs

I believe we should understand the components (i.e. heat, work, enthalpy,) because we will be doing problems based off of bomb calorimeter situations.

maldonadojs

Delta U insinuates you will be calculating the change in internal energy. Q in this equation represents the gained or lost heat and W represents the work being done on the system or work being done by.

maldonadojs

Heat is the measurable unit of transferable energy, while temperature is just a measurement of how fast the molecules are moving (i.e. this supports the idea of how gaseous molecules move much faster than those of a solid.

maldonadojs

I do not believe that there is a difference, but it follows the information given by the problem.

maldonadojs

Q is calculated whenever the equation is not at equilibrium, so if there is a change or anything similar. Also, Q is calculated whenever you are trying to see the result of the reaction (i.e. if the reaction favors the reactants or products).

maldonadojs

Yes that is the basics. The two strategies are to either flip the equations or multiply them by whole numbers. If you do decide to flip the equation you must make sure that you change the sign of the enthalpy. (I believe that is how you spell it)

maldonadojs

The only time you use the coefficients is when you’re trying to 1. solve an ice box and 2. using K equation.
K=[P]^coefficient/[R]^coefficient

maldonadojs

Conceptually they are different, for Kp is utilised for partial pressure and Kc for concentration. However, solving them is the same method.

maldonadojs

You definitely can use an ICE table for partial pressure as well. You would solve the same exact way, just the concept is a little different. For instance, the unit used is atp rather than mol/L. However, you would still set it up and Kp=Pproduct/Preactants.

maldonadojs

In this case, you would use the equilibrium concentration because the initial insinuates that the reverse reaction has not happened yet. So we are calculating what the concentrations when equilibrium will return.

maldonadojs

How does one calculate the 'C' value? And the 'C' does stand for change, correct?

maldonadojs

I enrolled a few days late and I am trying to catch up on the lecture notes. I am confused as to what the difference between Kc and Kp is. What is the difference and how am I supposed to acknowledge the differences.

maldonadojs

Every molecule with this formula will be some variation of square pyramidal. The bond angles of square pyramidal molecules are typically <90 degrees due to the fact that there is a lone pair of electrons that have the greatest repulsive force.

maldonadojs

Due to the fact that when concentration of [H+] increases, the substance becoming more acidic, the pH will be affected In the aspect that it will decrease.

maldonadojs

Resonance is, in the most basic way, a way to say that there are more than one way to draw a lewis structure. For instance, if the compound were to have more than on oxygen and one of the many had a double bond, you could draw the lewis structure with the double bond on a different oxygen each drawi...

maldonadojs

Inter- insinuates that the bond is between two molecules, while intra- insinuates that the bond is typically within the molecule between the separate atoms. The inter- forces are typically those of dipole-dipole etc. while the intra- are typically within covalent bonds.

maldonadojs

You use such prefixes when there is more than one polydentate ligands. So if there was two (en), you would use bis-prior to writing the name of the ligand.

maldonadojs

Distortion is a trend that happens when a highly polarizable atom is affected by a polarizing atom. The attraction results in the electron cloud to be shifted and misshaped. Also, if one is "highly polarizable", it is more likely to be distorted and affected by another atom.

maldonadojs

At this point when they all look similar, it is important to locate and count the amount of lone pairs. This is because of repulsion and attraction rules. The fact that one molecule may have lone pairs affects the connection between elements inside the molecule.

maldonadojs

From my understanding, the fact that the bond is covalent, the bonds must be touching. They share a pair of electrons, so that means we do assume their is such space between the two.

maldonadojs

Also, a way to check if the number of orbitals you have are correct, you can use the formula n^2. The n is the energy level. And to check if you need the number of electrons, you can use the formula 2n^2. It is more important to understand the concept and the n,l,m levels.

maldonadojs

It has to do with the fact that not everything is black or white. because some elements have more electrons, they are much more spread out and easier to connect with other elements.

maldonadojs

I may be definitely wrong but I thought that degeneracy meant same energy level, do it would be 4s?

maldonadojs

Due to the two being inversely proportional and that when you are sure of one, the other one is immeasurable.

maldonadojs

Today we discussed the basic chemical binding rules, as well as the basic Lewis structure rules. And we touched on polyatomic atomic species Lewis structures.

maldonadojs

You have to look at the periodic table and see which group the element is in and based off of which group, they have specific numbers of valence electrons. For instance, group one has 1 valence electron, while 2 has 2 and so on. Then based off of how many of that element is in the compound, you mult...

maldonadojs

The way I thought of it was that the energy that you find using this equation is the amount of energy that the electron gives off in order to return to a previous energy level. This is why the number is typically negative as well. The energy is the amount the it gives off say from the 6th energy lev...

maldonadojs

I am aware that they use the diameter because the electron is typically found outside of the diameter. In problems I have done I have noticed that the diameter was used to compare to the range of x values after solving the equation. Kind of a way to show that the electrons range even outside of the ...

maldonadojs

I was informed that whenever we are given a mass and are trying to find the wavelength, we must use De Broglie's equation. However, to answer your question, they do have wavelengths, but they are just too small for us to analyze. They can be as small as 10^-50 which is way past our capability. For i...

maldonadojs

The negative in this equation is due to the idea of the answer being energy lost. This is also the same amount of energy that a photon has in this situation. A photon in the experiment would have the same amount of energy as the energy lost from the electron.

maldonadojs

Lavelle also discussed in class that the answers will usually be easy to distinguish. For instance, the answers will be about 10^-12 rather than 10^20. He will not make it to where we have to use our own judgement to see if it is one thing or the other. He said they will be obvious enough for us to ...

maldonadojs

From my knowledge, the threshold energy is the minimum amount of energy it takes to remove one electron from the given metal, for each metal is different and requires a different threshold energy.

maldonadojs

I began this problem writing Magnesium Sulfate as MgSo4. Then the prefix hepta- refers to the number 7. Hydrate also refers to just water so it would result in MgSo4 7H2O.

maldonadojs

From my experience I have been told it is based off of the experiment, for some are notorious for having low % yield and some high. It depends because some experiments' procedures have more impurities or are more difficult and require more intense care. A short answer is no due to every experiment b...

maldonadojs

Would it be any different if I were to add their given masses and then divide by their combined molar masses in one mole and then divide by the volume? Or is it necessary to perform each mole calculation separately?

maldonadojs

When finding the answer for 'a', would that be considered using the M1V1=M2V2 equation? I am just trying to conceptualize this process.

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