CHEMISTRY COMMUNITY

VincentLe_3A

The review sessions will definitely be worth your time as the UAs are very good at helping you understand the concepts. In addition, going through the textbook problems and Sapling hw problems were also very helpful to me for past exams.

VincentLe_3A

F would be an electrophile as it would need one more to get a full octet. On the other hand, F- is nucleophile as it has an extra electron.

VincentLe_3A

According to the differential rate law, the relationship follows that the rate of the reaction is equal to the reciprocal of the stoichiometric coefficient times the instantaneous rate of change of the substance. Since No2 would have a coefficient of 2, so 1/a d[NO2]/dt would be 1/2 d[NO2]/dt where ...

VincentLe_3A

A catalyst would not show up in the overall reaction, as they are often a product of the first step of a reaction and product of a second step. Therefore, they would cancel out and the catalyst is not shown in the overall equation.

VincentLe_3A

A catalyst in a reaction mechanism often appears as a reactant in the first step and a product of the final step. An intermediate on the other hand shows up as a product of the first step and reactant of the second step. Both are not shown in the overall reaction equation.

VincentLe_3A

The e naught signifies reduction potential at standard conditions, whereas e is reduction potential with other conditions.

VincentLe_3A

The only factor which affects k is a change in temperature.

VincentLe_3A

The main function of the salt bridge is to prevent charge buildup within the galvanic cell. It does this by transferring ions to keep the solution neutral, allowing electrons to continue flowing from anode to cathode.

VincentLe_3A

First order reaction can have more than one reactant as long as it fulfills the condition where one of the reactants affects the order of the reaction while the others do not. In other words, one reactant must be to the power of 1 while the others can only be to the zero power.

VincentLe_3A

The 0.693 comes from ln(2) when simplifying the equation.

VincentLe_3A

Yes, when a battery is dead, there is no electrical potential difference between the two cells. Therefore, there is an E potential of 0, which makes delta G zero.

VincentLe_3A

Yes, when E naught Cell is positive, then the reaction is spontaneous.

VincentLe_3A

The general rule is that when K>1, meaning the amount of products is greater than reactants at equilibrium, then the reaction is favored in the forward direction and therefore, spontaneous, making delta g<0. When K<1, the amount of reactants is greater than products at equilibrium, the reaction is n...

VincentLe_3A

E naught is the cell potential under standard conditions of 1 atm, 298K, and 1 M. E is the cell potential when the conditions are altered from the standard conditions, such as a change in temperature or some other parameter.

VincentLe_3A

n represents the number of moles of electrons transferred within the redox reaction.

VincentLe_3A

It is important to first determine and define the system in the scenario. Then, once the system is defined, then work is negative when the system does work. If work is done on the system, then the sign of work would be positive.

VincentLe_3A

Delta G naught is the change in Gibbs free energy at standard conditions, whereas Delta G is the change at a specific condition. The key to determining which to use is to see if standard condition values are given or if the system is at equilibrium, then that means Delta G is zero.

VincentLe_3A

Adiabatic is a term which refers to a system where there is no heat exchange between the system and its surroundings. Specifically, this would mean q=0.

VincentLe_3A

Work of compression is negative based on what you decide as the system. We usually refer to the gas inside the piston as the system, so in that case, work is being done on the system when compressed, so work of compression would be positive on that system. However, if we define the system as the sur...

VincentLe_3A

You would utilize the Cv constant in systems under conditions of constant volume while Cp in systems under conditions of constant pressure.

VincentLe_3A

Yes in the term q(rev), the "rev" refers to a reversible process. Therefore, q(rev) is the heat of a reversible system.

VincentLe_3A

Yes, you can equate delta s total to delta s universe because that is the total entropy in the universe. The way I think of it is that the total entropy is the same as all entropy in the universe.

VincentLe_3A

I believe the T is referring to the temperature of the surroundings once the reaction in completely done.

VincentLe_3A

In determining which value of R to use, it is important to consider the units of the other given values. You must make sure the units cancel accordingly, so you must use the R value which allows for unit cancellations.

VincentLe_3A

ΔG°is Gibbs Free Energy under standard conditions, whereas ΔG is Gibbs Free Energy based on conditions given in the problem.

VincentLe_3A

When a system does work, there is a negative value for work because you can think of it as energy being inputted (and thereby released) by the system. The surroundings of the system would then absorb that energy inputted. However, when work is done to a system, the system itself is essentially absor...

VincentLe_3A

I believe it is important to have a conceptual understanding of how different calorimeters work in order to determine whether the system is open or closed. This determination will affect the way we approach a problem and thus lead to different calculations.

VincentLe_3A

An inert gas is a gas which does not react with the reactants or products of the chemical reaction in a certain system. Therefore, since there is no interaction, the inert gas would not affect the ratios of products and reactants, thus leaving the equilibrium constant unchanged.

VincentLe_3A

Personally, I believe going through the textbook problems as well as attending UA sessions are the most effective methods to study for this class. UA sessions are really helpful in going over the material from a previous student's perspective.

VincentLe_3A

Yes, Dr. Lavelle said in lecture that residual entropy is the same thing as positional/statistical entropy. They are interchangeable.

VincentLe_3A

Yes, I believe the 5% rule applies to both molar concentrations as well as partial pressures. The reason being is that the conversion between the two is a simple calculation using the gas constant and temperature. Assuming that the temperature stays the same, there would be the division or multiplic...

VincentLe_3A

Condensation would be considered an exothermic reaction as you are going from a gas to a liquid. Another way to think of it is that this process forms bonds, and the process of forming bonds release energy, thereby being exothermic.

VincentLe_3A

The ratio of (Po2)/(Po3) would not stay the same because once the 0.5 M of o3 was added, there would be a change in the individual partial pressures of both Po2 and Po3. Because the ratio (Po2)/(Po3) is not the equilibrium constant, it would not stay the same, as the molar ratios from the stoichiome...

VincentLe_3A

According to the lectures, it is safe to approximate when the equilibrium constant K is less than 10^-4. To be sure that approximation is appropriate, always check for the 5% rule afterwards, where if the percent ionization is less than 5%, then approximation is appropriate.

VincentLe_3A

I do not believe we have to memorize bond enthalpies since there are so many different values. Also, the homework problems had us refer to the table, so I assume we are given them.

VincentLe_3A

The equilibrium constant essentially tells us the ratio of products to reactants at equilibrium. This is useful in determining many things if asked about the system. For instance, we can compare this value to a reaction quotient Q to K to determine which direction the reaction will shift to in order...

VincentLe_3A

In the specific direction that the reaction takes place, the reactants would have negative change because they are being used up while the products would have a positive change because they are being formed. The coefficients for these changes can be found in corresponding stoichiometric coefficients...

VincentLe_3A

We only include water in equilibrium constant K calculations when water is not in excess as a solvent. Essentially, this means only include water in K calculations when it is in its gaseous phase with a partial pressure, where it does not act as a solvent in the system.

VincentLe_3A

Yes, I believe both terms mean the same thing since it is in the same phase.

VincentLe_3A

K would not change with the addition or removal of reactants or products, since such addition or removal of molecules will make the system re-establish the ratio of K. The only time K changes is in the case of change in temperature.

VincentLe_3A

Only gases are affected by an increase or decrease in pressure through a change in volume specifically. It is important to note that a change in pressure due to an addition of an inert gas will have no effect on the overall reaction.

VincentLe_3A

I believe the two statements can be used interchangeably. When it "favors" or "shifts" right, then products are formed, but when it "favors" or shifts" left, then reactants are formed.

VincentLe_3A

We would first have to determine if the reaction given is endothermic or exothermic, which can be seen either through a positive or negative value of Delta H. A positive Delta H would indicate an endothermic reaction, and an increase in temperature for an endothermic reaction will favor products. A ...

VincentLe_3A

Yes, q will eventually reach k after letting the reaction take place over time. Q is finding the ratio at a certain time during the reaction, which could be greater than, less than, or equal to k. If the q is equal to k, then the reaction is at equilibrium. However, if q<k, that means there are more...

VincentLe_3A

In ICE problems where the equilibrium constant is Kc, which is specifically for the concentrations of each species, we must find the concentrations in Molarity for each species. Therefore, he divides 1.5 moles by 0.5 L to find the concentration of the species in terms of moles per liter.

VincentLe_3A

You can determine which gas constant R to use based on the units given in the problem. The units should cancel out for whichever variable you are trying to determine.

VincentLe_3A

As stated by Le Chatelier's Principle, chemical reactions will adjust accordingly in order to minimize the effect of changes. Therefore, if the concentration of the products is decreased, the reaction will proceed towards the product side in order to replace the products that were displaced. The opp...

VincentLe_3A

An ideal gas is a conceptual idea of a gas in which the gas particles do not experience intermolecular forces or lose energy during collisions. The ideal gas law equation can be useful in calculations, such as converting between partial pressure and molar concentrations.

VincentLe_3A

The equilibrium constant K does not change as long as the temperature is held constant. We compare Q, the reaction quotient which is a certain ratio at a certain time, to the K, which is the ratio at the reaction equilibrium. When Q<K, this indicates more reactants than products compared to the equi...

VincentLe_3A

Yes, a negative delta H indicates that energy is released on the product side, thereby being an exothermic reaction. A positive delta H indicates that energy is absorbed on the reactant side of the reaction, making it an endothermic reaction.

VincentLe_3A

Delocalization of electrons means that the electron density will be better evenly spread out throughout the molecule instead of being pulled to one particular area. This allows for more stability, as it would be less reactive with other charged molecules if the electron density more evenly distribut...

VincentLe_3A

Yes, alkaline solutions are referring to basic solutions, which have a pH>7 and have a higher hydroxide concentration than hydronium concentration.

VincentLe_3A

If you take a look at the manipulated version of the Heisenberg Uncertainty equation, where we change Δp to mΔv and divide it on both sides, we get Δx >= h/ (4π)(mΔv). According to this equation, mass is inversely proportional to the uncertainty in position. So a really large mass would result in a ...

VincentLe_3A

I had a similar question, and I just want to check my understanding. So electron geometry just takes into account areas of electron density while molecular shape takes into consideration whether those areas are an atom or a lone pair. Will we be asked to find electron geometry or molecular shape? H...

VincentLe_3A

I believe the format will be similar to our previous exams, so only some of the test questions will be pulled from the homework.

VincentLe_3A

Yes, an alkaline solution is referring to a basic solution, meaning it has a pH>7 and contains more hydroxide ions than hydronium ions,

VincentLe_3A

The charge of the transition metal cobalt would have to be +3 because the overall charge of the coordination compound would have to be +1 in order to compensate for the -1 charge of the chloride anion. Since there are 2 chlorine anions within the coordination compound, the charge of the cobalt would...

VincentLe_3A

If you visualize 6 bonds to a central atom, there would be 4 equatorial bonds and 2 axial bonds. There would be 8 triangular "faces" if you were to connect all of these bonds to visualize the shape. Therefore, it would be octahedral because of these 8 faces, which resemble a octahedron sha...

VincentLe_3A

The oxidation state is the number which represents the charge and number of electrons that an atom lost or gained. A positive oxidation number indicates loss of electrons as the atom is more positively charged while a negative oxidation number indicates a gain in electrons since the atom would be mo...

VincentLe_3A

Coordination number is the number of bonds on the transition metal cation within the coordination compound. It is important to note that a single ligand can contribute to more than one coordination number if it is polydentate. For instance, a bidentate ligand, such as ethylenediamine, would be able ...

VincentLe_3A

So our transition metal is Co and the ligands are (en) and CO. CO is a monodentate ligand, so since there are two of those ligands, each one forms a single bond with the transition metal, adding 2 to the coordination number. However, (en) is a bidentate ligand, meaning a single (en) will form two bo...

VincentLe_3A

Finding hybridization involves counting the number of regions of electron density around the atom. In the carbon, it has 4 single bonds, 3 to each hydrogen and 1 to the oxygen. This results in a sp3 hybridization as there are 4 regions of electron density. Oxygen has two lone pairs of electrons, a b...

VincentLe_3A

The polarity of a molecule depends on both its shape and dipole moments between the individual atoms. The most methodical way of finding polarity would be to draw out the lewis structure and figure out the VSEPR shape. From there, draw out the arrows which represent the dipole moments in which the e...

VincentLe_3A

The molecular formula suggests the bonding structure between the atoms within the molecule. Especially in molecules with carbon and hydrogen, carbon is the backbone of the molecule, so make the bonds between carbons, add in the hydrogens, and finally place double bonds where needed to fulfill octet ...

VincentLe_3A

The polarity of a molecule depends on both the shape and dipole moments. Drawing out the molecule is the best way to get a full visual representation, which will allow you to see if the dipoles cancel out. The shape does contribute to whether the molecule is polar or non polar, but sometimes molecul...

VincentLe_3A

Just as everyone else has said, going through the homework and Sapling questions is a very helpful tool when studying, as those are similar to the test questions on the exams. In addition to that, I like to take a sheet of paper and condense all my notes with the material that is going to be on the ...

VincentLe_3A

Since the energy of 3d orbital becomes lower than that of the 4s orbital once it starts getting filled with electrons, it makes sense to write 3d before 4s in terms of energy order of the orbital. It is important to note that the 4s orbital is lower energy than the 3d orbital until the 3d orbital st...

VincentLe_3A

The general guideline is that most elements in period 3 or higher in the p-block can have an expanded octant because of the d-orbitals in the valence shell of those elements which can accommodate additional electrons. I believe they do not necessarily have to be empty for the element to have an expa...

VincentLe_3A

Molecules with more covalent character have a smaller electronegativity difference compared to molecules with ionic character. As a result of this smaller difference, there is no one atom which completely takes all the electrons, thereby making it more covalent. The important thing to consider when ...

VincentLe_3A

A coordinate covalent bond is a bond in which a shared pair of electrons originates from one of the atoms. In a typical covalent bond, one electron from one atom is shared with one electron from another atom. However, in a coordinate covalent bond, the electrons come from one of the two atoms in the...

VincentLe_3A

The most important thing to consider when determining ionic character within a molecule is the difference in electronegativity between the atoms. Molecule size has some relation towards this, as the general trend is that smaller atoms have higher electronegativity. Therefore, generally a hydrogen bo...

VincentLe_3A

I believe because the problem is asking which structures are most plausible based on formal charges, choices C and D are the only ones out of the given structures which align with the charge of -1 in the overall molecule. Given that the molecule has resonance, these two structures are possible repre...

VincentLe_3A

For hydrogen bonding to occur, there must be an attraction between a slightly positive hydrogen and a nitrogen, oxygen, or fluorine with lone pairs. In the diagram shown, there are 4 hydrogen atoms bonded to 2 nitrogen atoms with lone pairs of electrons. The hydrogen atoms are slightly negative, mak...

VincentLe_3A

Following the guidelines for coming up with electron configurations, we would think Cr would have an electron configuration of [Ar] 3d^4 4s^2. However, in the 3d subshell, it is more stable if all 5 orbitals in that subshell have an electron. Therefore, instead of 4s^2 having a pair of electrons in ...

VincentLe_3A

Lewis acids are electron pair acceptors, meaning they are able to attract a pair of electrons. Usually these are electron deficient cations or an atom or molecule that has an incomplete octet. On the other hand, Lewis bases are electron pair donors, meaning they are able to donate a pair of electron...

VincentLe_3A

Since he drew the arrow in the direction of the partially negative atom in lecture, I believe that is what he expects us to use. Conceptually, I believe this makes more sense than the arrow pointing towards the partially positive because the electron density would be greater towards the partially ne...

VincentLe_3A

Electrons in the p-orbital are further away from the nucleus compared to those in the s-orbital. Therefore, since they are closer to nucleus, the inner electrons in the s-orbital shield the nuclear charge from the outer atoms in the p-orbital, since the attraction is more directed towards the closer...

VincentLe_3A

For me personally, I find myself often tired and unfocused in my bedroom, so what I do is find a spot in the house which I use exclusively for studying. This change in environment is beneficial because it conditions my brain to think that it is "study time" whenever I go to this spot, inst...

VincentLe_3A

I believe it is just conceptually important to understand that double bonds are shorter than single bonds, but we do not have to visually show it on the Lewis structures because the main focus of the Lewis structures is the valence electrons and the relationship of the valence electrons between diff...

VincentLe_3A

To add on, since the electrons are delocalized, the amount of electron-electron repulsion within the structure is decreased, thus allowing it to be more stable than if the electrons were not delocalized.

VincentLe_3A

If possible, it is ideal for the formal charges of the atoms to be zero, as it indicates the most stable state. However, given certain situations, some molecules may not be able to have everything with a formal charge of zero. In that case, it is just important to remember that all the formal charge...

VincentLe_3A

To solve these problems, you look at the specificity of the quantum numbers given. For example, in the first problem you are given n=3, so you know it is talking about the 3rd energy shell. In the third shell, you have s, p, and d orbitals (you can figure this out by finding the allowed values of l,...

VincentLe_3A

For the first problem with n=4 and l=2, you must consider that this is specifically the 4d orbital, which does not include the 4s or 4p orbitals. I think you are grouping all these orbitals together, which is incorrect because electrons in the 4s would have the quantum numbers of n=4 and l=0 while e...

VincentLe_3A

I believe your answers are correct. You can try using only 3 significant figures (ex. Proton= 0.689 m/s) instead of 4 significant figures because it showed up as correct for me.

VincentLe_3A

Oxygen has a lower first ionization energy compared to nitrogen because when we look at their electron configuration in the p subshell, we see that nitrogen has three unpaired electrons in each of the px, py, and pz orbitals. However, oxygen has one pair of electrons in one of those orbitals in the ...

VincentLe_3A

For the Rydberg equation which you are using, it states that the frequency (v) = R((1/n1^2)-(1/n2^2)). For the equation you have right now, you are making it equal to the wavelength, so use v (frequency)= λ/c to find the frequency. Then from there, plug the frequency into the Rydberg equation and yo...

VincentLe_3A

After figuring out all the other missing information, the main aspect that will lead to knowing the event is the wavelength of the wave. In order to determine the event, you must be familiar with the electromagnetic spectrum and the different regions for certain wavelengths (Radio waves, Infrared, V...

VincentLe_3A

The Balmer series corresponds with visible light because the wavelength of the electromagnetic radiation that was absorbed/emitted had a wavelength in the visible light region (within the range of approximately 400-700 nm). It is the same for the Lyman series: the electromagnetic radiation that was ...

VincentLe_3A

When an electromagnetic emission problem mentions a radiation of a wavelength in the Lyman series, it means the electron goes from some higher energy level (n>=2) down to the first energy level (n=1), so for the Lyman series, the final electron energy level is n=1. For the Balmer series, it means th...

VincentLe_3A

I think either equation is acceptable to use on the midterm, and especially since the midterm is timed, it would make sense to use the Rydberg equation frequency=R(1/(n1)^2 - 1/(n2)^2) in order to save time. As long as you understand conceptually where that equation comes from, the Rydberg equation ...

VincentLe_3A

I think it is just important to know the overall order, but not specifically the different wavelengths for each other than the range for visible light. One way I remember the spectrum is that from the red visible light (around 700 nm), it is followed by infra"red" which is longer. Then for...

VincentLe_3A

To add on, it is just conceptually important to distinguish the difference between the classical mechanics model and quantum mechanics model of light. In the classic mechanics model, light acts as a wave whereas in the quantum mechanics model, light is absorbed and emitted in discrete units (photons...

VincentLe_3A

To add on to what the other people have said, it is important to note that the kinetic energy of the electron will only be present if the energy of the photon exceeds the work function (threshold energy). As seen in the equation Energy of Photon - Work Function= kinetic energy of electron, if the en...

VincentLe_3A

For the two equations E=hv and C= lambda(v), they have the common term "v" for frequency. To isolate the variable "v" in the equation C=lambda(v), you would divide lambda on both sides. This will give you v= c/lambda. From there, you would plug in v= c/lambda into the variable &q...

VincentLe_3A

Hi, so Euler's number, which is often represented as a lowercase e, does have a different value than 10. However, this is not to be confused with the one shown on your calculator, which is an uppercase E that refers to "exponent" with a base of 10 followed by a number to represent the powe...

VincentLe_3A

When referring to light using the classical wave model, an increase in the amplitude of the light wave would lead to an increase in the intensity (and vice versa). However, when we are considering light through quantum mechanics, an increase in intensity would mean an increase in the number of photo...

VincentLe_3A

When a problem tells you that a certain molecule reacts with an excess of another molecule, does that mean that the excess of the other molecule is one of the reactants? For example on the last two questions on the Sapling Homework, the problems stated (9)"Caproic Acid is burned in excess oxyge...

VincentLe_3A

This problem gives you the radius of each Ag atom, which is 144 pm. You would have to figure out the number of atoms in the given amount of Ag atoms. Since 1 mole of Ag atoms was given, you can just multiply Avogadro's constant (since 1 mole= 6.0221* 10^23) by 288 pm (which is the diameter of each a...

VincentLe_3A

The moles of the solute will stay the same throughout dilution since we are not adding or removing any amounts of the solute. Dilution only changes the concentration of the solute within a certain volume, not the amount of solute itself.

VincentLe_3A

We use Avogadro's number in questions that ask specifically for the "number of objects". For instance, if the problem asks for the number of atoms/particles in an element or molecules in a molecular compound, then you would use Avogadro's number. If the problem gives you the amount in mole...

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