CHEMISTRY COMMUNITY

Kyle Dizon 3A

The question should provide the information needed to identify which units is appropriate to use, but this is usually calculated in seconds.

Kyle Dizon 3A

I think an intermediate is a species that is produced but later consumed again in the reaction.

Kyle Dizon 3A

Catalyst is not considered a reactant or a product because it will not appear in the end of the overall reaction.

Kyle Dizon 3A

Catalyst are just used to help speed up a reaction but they are not necessarily consumed. It is not considered a reactant or a product.

Kyle Dizon 3A

The units would be kj/mol or j/mol.

Kyle Dizon 3A

If the temperature is in 25 degrees celsius or 298 kelvin, then you must use the equation that has the log. Everything else will be using the equation that consists of ln.

Kyle Dizon 3A

The stoichiometric coefficient is what determines a. In the lecture, it just so happens to be that the coefficient 1.

Kyle Dizon 3A

A change in K ( temperature ) or a catalyst will alter or affect the rate constant.

Kyle Dizon 3A

Temperature is the only factor that has the ability to affect K.

Kyle Dizon 3A

Our professor mentioned that this is possible but will not be covered or given anything beyond third order reaction.

Kyle Dizon 3A

You need a metal conductor in the reaction if there are none. So, you would add Pt to conduct electrons.

Kyle Dizon 3A

You are able to obtain the moles from the balanced redox reaction to find delta G.

Kyle Dizon 3A

Yes. The cathode is located on the right side while the anode is supposed to be on the left side.

Kyle Dizon 3A

Yes, the reduced molecule is the oxidizing agent of the system.

Kyle Dizon 3A

Personally, I have always used the acronym of OIL RIG when I learned it in high school. When oxidation occurs, you lose electrons while reduction is gaining the electrons.

Kyle Dizon 3A

STP refers to 0 celsius which converting this becomes 273.15 kelvins.

Kyle Dizon 3A

The only way I am surviving the calculations right now is referring to the units. Use the equation sheet to refer to the proper units to identify the correct steps to get the units needed.

Kyle Dizon 3A

We learned this back in the first few weeks, but if Q is larger than K, the products would be favored. Having a smaller Q will result in a favor of reactants. If they are equal, they are claimed to be in equilibrium.

Kyle Dizon 3A

Delta G simply just refers to the free energy of the system. The delta G naught often refers to a more specific standard conditions of free energy.

Kyle Dizon 3A

Possibly using kelvin will be the most reliable answer instead of celsius. If you are given celsius, simply convert to kelvin by adding 273. It is also important to remember to use the proper units for the R constant.

Kyle Dizon 3A

The reversible reaction is presented by qrev, while q just represents the heat.

Kyle Dizon 3A

The symbol refers to the standard free energy of a system, specifically its conditions under 298 Kelvin and 1 atm.

Kyle Dizon 3A

Molecules that are large will have more room to occupy different positions, in this case disorders, therefore the entropy value becomes higher.

Kyle Dizon 3A

In this regards, this is explaining the relationship of the the reaction and calorimeter in which it says that the heat of the calorimeter is the same as the heat of the reaction.

Kyle Dizon 3A

The heat identifies if the system is endothermic or exothermic, but work does not relate with this concept.

Kyle Dizon 3A

It would possibly help to put yourself in a position where you are the system to determine the sign for work. Imagine yourself in a hot room where you start to fan yourself to help yourself cool down. Once you begin fanning, you are using energy meaning you are doing work so you lose energy and work...

Kyle Dizon 3A

The first law of thermodynamics is the general idea of conservation of energy, In other words, energy cannot be destroyed or created. As explained, energy states that if energy is released, it is absorbed by another substance.

Kyle Dizon 3A

From how I understand it, you can think of yourself as riding a bicycle. In order to achieve greater speeds, you would have to pedal meaning, you are doing work therefore losing energy (-w). Then, when work is done for you when you are going downhill (surroundings) you gain energy. Hope this helps!

Kyle Dizon 3A

Yes, q will always be zero in adiabatic systems due to the fact that there are no heat transfers released.

Kyle Dizon 3A

If there was a system, the external pressure would be the surroundings outside of that system. In this case, it would be the pressure that is caused by that piston and the internal pressure would be the pressure pushing against that piston if it was expanding.

Kyle Dizon 3A

It could be similar to vacuum but to an extent. It would be more reasonable to use a bomb calorimeter to describe an isolated system as the outside environment will not affect the system.

Kyle Dizon 3A

Right, an isolated system is protected from outside environment making the values 0.

Kyle Dizon 3A

A perfect system would be that the heat that is exerted from the reaction is equal to by the absorption of the solution that ultimately means all energy is still present.

Kyle Dizon 3A

Molar heat capacity can be mostly found in the textbook problems we are assigned in the outlines as the textbook explains that it is the amount of heat required for raising a mole of a substance.

Kyle Dizon 3A

An enthalpy is a state function which allows its properties to be added or subtracted.

Kyle Dizon 3A

To begin, state properties are properties that can be identified through the final and initial value such as enthalpies, energy, etc. The work to achieve a final value is not considered in state properties.

Kyle Dizon 3A

To find out how a reaction reacts from a change in partial pressure, you must identify if the pressure increases or decreases. Once this is identified, you will be able to identify if the reaction shifts forward or backwards to reach in order to reach equilibrium.

Kyle Dizon 3A

From what I remember during lecture, Le Chatelier's explains this situation in which if there is a change in the pressure or concentration of a reactant or product, it affects the equilibrium. The reaction just then shifts until it can reach equilibrium once again.

Kyle Dizon 3A

Yes they are all zero as said in the lecture. From what I remember, he went through all seven of them when he pointed to the periodic table during lecture.

Kyle Dizon 3A

Yes the units will always be kj/mol unless it is the ΔH⊖rxn because that is the overall reaction of enthalpy.

Kyle Dizon 3A

If this was to happen, the value of K would also be changed due to the temperature change. Additionally, this would require extra information such as finding two values of Q that are different.

Kyle Dizon 3A

[H2O] will be left out due to the large excess which is why it can be disregarded due to its presence in both the reactant and products as stated in the lecture.

Kyle Dizon 3A

Based on some of the UA's, an equilibrium constant smaller than 10^-4 would allow you to disregard the x for an approximation.

Kyle Dizon 3A

I believe that the a and b are simply to label and distinguish acid and bases. Ka representing the dissociation of acid while Kb deals with base which provides the Kw value.

Kyle Dizon 3A

Solid and liquids are omitted due to the fact that they do not have concentrations. Therefore, they would have no influence on the K and will not favor reactants or products.

Kyle Dizon 3A

K, the equilibrium constant can only be affected if the temperature is altered. Other factors do not affect this as discussed in the previous lecture that the constant will still be the same due to the ratio. Changing factors like pressure will simply change when equilibrium is reached.

Kyle Dizon 3A

Q and K are found using the same concept of Product/Reactants. The main difference between the two is that the reaction quotient (Q) is identified when the reaction is not at equilibrium while K is the actual equilibrium constant. Comparing these two will provide us information if the reactant or pr...

Kyle Dizon 3A

Correct. Product divided by reactants will determine the K. If reaction is not at equilibrium, Q can be determined in order to identify if the reactant or products will be favored.

Kyle Dizon 3A

Right. Temperature is the only factor that could change the K due to its affect during the chemical reaction.

Kyle Dizon 3A

Yes, we often use Q in order to determine if the reactant or product will be favored.

Kyle Dizon 3A

To add on I have always approached acids and bases by creating my theory that lewis acids are electron deprived and bases are electron rich that can donate to these acids.

Kyle Dizon 3A

Lewis acids are often electron acceptors because they need the electron from lone pairs to satisfy the valence electrons that they need. From how I see it, lewis base are those who can afford donating or sharing the lone pairs around the orbitals that they have.

Kyle Dizon 3A

The way I understood it is to understand that Lewis Acids need the electron that makes them the acceptor because their orbitals or “cloud” is empty. The Lewis Bases have electrons in their cloud that they are able to donate once there is bond that is made.

Kyle Dizon 3A

From what I recall that I got from the lecture, both of these have similarly concepts, only that Bronsteds are associated with the protons acceptor while the Lewis bases are associated with the electron donations. I believe he said that they are both similar principles so in regards of defining, we ...

Kyle Dizon 3A

Water is able to act both as an acid or a base.

Kyle Dizon 3A

Lone pairs influence the bond angles by making them decrease because of the repulsion that it causes. In regards of knowing the actual bond angle, we do not need to know the exact angle, only the general idea of how it is influenced.

Kyle Dizon 3A

Analyzing the shape will help provide a clear understanding of what the bond angles could possibly be. Directly answering your question, yes, a linear molecular shape has a bond angle of 180 degrees.

Kyle Dizon 3A

I think memorizing that the lone pairs and bond pairs influence the bond angles of the molecular shape/geometry. The way I memorize it is that when a lone pair present, the repulsion has a larger effect causing the bond pairs to be pushed more closely together.

Kyle Dizon 3A

Emphasizing what the others have said, it is probably better to just memorize it. The resources that is provided by the TA’s and UA’s will probably be our best bet in regards of being familiar with the shapes, bond angles, etc.

Kyle Dizon 3A

I am unsure but I do not recall the d orbitals being discussed or taught during his lecture, but the s and p orbitals will be on the final exam.

Kyle Dizon 3A

It would be a trigonal pyramid due to information given of the three oxygen bonded with the sulfur along with one lone pair.

Kyle Dizon 3A

The angles are experimentally determined and can not be calculated precisely. In this case, the VSEPR will help us the most in determining which angles are larger or smaller based on the bonds that are provided for us to idenitfy.

Kyle Dizon 3A

From the knowledge I have gained so far, I believe bond angles are determined experimentally. Bond angles change based on the electron repulsion on the different bonds that are given to us.

Kyle Dizon 3A

It is probably essential that we know all of the shapes that was discussed during the lecture. Just to be safe, I think it will be better if we familiarize ourselves beyond what we need to know to prepare us for organic chemistry.

Kyle Dizon 3A

Personally, the only way that it has been helping me is to hop on quizlet and memorize from there.

Kyle Dizon 3A

I think it was written this way because the orbitals that are found in 5s are the outermost shell which means that it easier to take away electrons rather than the 4d shell which it is closer to the nucleus.

Kyle Dizon 3A

From my knowledge, the electronegativity is determined experimentally which means that we would be provided values and a table to identify what the question is asking for. Also, even if we do not need find it experimentally, the trends in the periodic table should provide sufficient information to f...

Kyle Dizon 3A

Wondering the same thing honestly. I think as long as we are familiar with the trends in the periodic table we should be fine on the midterms and exams. If needed, I think the tables will be provided for us if it is needed in the problem.

Kyle Dizon 3A

In CH4, LDR exists because of the fact that C and H do not have much of a difference in their electronegativity. Adding on, drawing the Lewis structure will help visualize that the hydrogen surrounding the atom in the middle are all similar causing the dipole moments to cancel and not have an effect...

Kyle Dizon 3A

In the lectures and other sessions, the information they provide has the arrow leaning towards the electronegative atom. I would say to ultimately just follow how the lecture demonstrates it rather than the book. In regards of understanding the concept better, I think just go with your personal pref...

Kyle Dizon 3A

Had the same issue regarding this problem! I figured it was an error in the problem itself but I said it was dipole-dipole.

Kyle Dizon 3A

The atom with the lowest ionization energy goes in the middle of the lewis structure because they are more compatible to accept bonds and electrons with other atoms.

Kyle Dizon 3A

Lone pairs are essentially electrons that belong to that certain atom that is being described. They are not bonded with any other elements.

Kyle Dizon 3A

The elements that you have listed essentially describes the octet rule exceptions in which they are able to accommodate extra electrons due to the period 3 / or 3d subshell.

Kyle Dizon 3A

Essentially, the electronegativity trend shows that it increases from left to right and bottom to top in the periodic table.

Kyle Dizon 3A

I do not think that the size of an atom is influenced by the electronegativity that it contains, but we do see that it is somewhat the opposite of each other because the periodic table shows a trend that electronegativity gradually increases when going left to right/bottom to top of the table. On th...

Kyle Dizon 3A

I think these are exceptions due to the fact that the valence electron bonding isn’t strong enough to hold all of its electrons.

Kyle Dizon 3A

Yes, lewis structures can be used to represent ionic bonds similar to the example that Professor Lavelle demonstrated in the previous lecture during week 4.

Kyle Dizon 3A

The octet rule does not apply to the elements of H, He, Li, and Be. While there may be more elements that do not apply to the octet rule like Boron, Professor has only discussed these four so far.

Kyle Dizon 3A

I am pretty sure it means that it is due Sunday at 11:59pm just like our five post in chem community.

Kyle Dizon 3A

Yes that is true, the octet rule applies to Boron because of the three valence electrons that it possesses.

Kyle Dizon 3A

They do not lose any electrons due to their ionization energy. The bonds are stronger because of they have more pull/attraction making them to be poor electric conductors.

Kyle Dizon 3A

Essentially, covalent bonds often share electrons between other elements while the ionic bonds simply take or give electrons off depending on the situation.

Kyle Dizon 3A

Hydrogen bonds are often have the weakest link while covalent bonds have the strongest link with each other. In addition, the bonds with single bonds are ideally weaker than triple bonds.

Kyle Dizon 3A

I believe the 3d would have to be filled up by an electron to its full capacity in order for the electron to go to the next energy level.

Kyle Dizon 3A

The photon would be emitted once the excited electron begins to come down from its peak energy level, once it reaches normal state there will no longer any emitting that occur because all the photons have been emitted.

Kyle Dizon 3A

The intensity is related to the number of photons of the light.

Kyle Dizon 3A

I would stick with the SI units just in case because it is easier to work with the SI units.

Kyle Dizon 3A

I think in regards of calculations, you will be fine if you are off by a few numbers. Best to refer to the answers to the multiple choice question if your calculation is slightly off.

Kyle Dizon 3A

It will be easier to remind yourself that it should be final minus the initial.

Kyle Dizon 3A

Our midterm will be all multiple choice, so I think everything we need in regards of equations will be provided and given to us on the equation and constants worksheet.

Kyle Dizon 3A

Often, I try to relax by changing the scenery where I am and surrounding myself with sounds that calm me down such as songs. Personally, I try to listen to my body and my mind in order to avoid over stressing and preventing anxiety attacks. Try to spread out your studying time as well.

Kyle Dizon 3A

I think you should master the equation just in case it will show up on the midterm. Since it was on the lecture, it is a possibility that it will be an equation that we may require to use during the midterm.

Kyle Dizon 3A

Based on my knowledge, we only use the Heisenberg indeterminacy equation once we are asked for the velocity.

Kyle Dizon 3A

Relying on the units will help with your confusion. Also, you can attempt to multiply the percentage to find the value of the uncertainty.

Kyle Dizon 3A

Energy is usually given in Joules after calculations. If you are calculating for wavelength, the problem will often tell you to either give your answer in meters or nanometers.

Kyle Dizon 3A

I believe it depends on what you are trying to calculate and what it is asking for. For the most part, it will probably ask for nanometers.

Kyle Dizon 3A

Essentially, I think it is all about time management. Creating a time frame throughout the day for each of your classes can help you stay on top them. For chem, I try to watch the lecture twice, the first time trying to focus on writing all the information that I need and the second to pay attention...

Kyle Dizon 3A

That equation is the method to find kinetic energy. Professor Lavelle mentions that this calculation expresses the excess energy of the electron that is emitted. Please correct me if I am wrong, hope this helps.

Kyle Dizon 3A

Based on the information I have gathered from my TA and Professor’s email, we do not necessarily need a webcam if we have another device that will show our workspace while taking the assessments.

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