CHEMISTRY COMMUNITY

Isha_Maniyar_Dis2E

I'm not entirely sure about this, but I believe you can't tell whether the energy in each step increases or decreases. It can happen either way, depending on the individual delta H of each step.

Hope this helped!

Isha_Maniyar_Dis2E

The rate is independent of [C] because when you look at rates 1 and 4, you can see that while the concentration of C stays the same, the rates still change. As a result, C is zero'th order. Set the concentration of either A or B as constant and compare 2 rates. See how the rate changes as concentrat...

Isha_Maniyar_Dis2E

To add on, adsorption is defined as reactants on the surface of the catalyst. For this to occur, I believe it is implied that the catalyst will have to be solid, so that there can be a surface for the reactants to rest on and react with. If it were a homogenous catalyst, the complex would likely for...

Isha_Maniyar_Dis2E

An increase of 0.034 in B will result in a 0.068 decrease in A, because A decreases at twice the rate of B increasing. You then subtract 0.068 from [A]₀ to get [A], which you can plug into your equation.

Hope this helped!

Isha_Maniyar_Dis2E

Use the ratio between A and B to figure out how much A will decrease by when B increases. You can then figure out [A] by subtracting this value from [A]₀.

Because B is 3 times the amount of A, A will decrease by 1/3 the rate of B. So, you divide B by 3.

Hope this helped!

Isha_Maniyar_Dis2E

Electrolytic cells push the cell in a non-spontaneous direction. Therefore, Ecell will be negative, and ∆G will be positive.

Hope this helped!

Isha_Maniyar_Dis2E

You have to look at the solubility constant, K_sp. This tells you how soluble a substance is, and whether it will dissolve in water.

Hope this helped!

Isha_Maniyar_Dis2E

K+ and Cl- are spectator ions, and because this is a net ionic equation, you do not need to include them in the reaction.

When you are writing your half-reactions, only include the the species that are being oxidized or reduced.

Hope this helped!

Isha_Maniyar_Dis2E

You know that K decreases as temperature increases. This means that the reaction is exothermic. Because of this, the reactants are favored, and increasing the temperature will result in a shift to the reactant side. More ammonia will NOT be formed, because this is a product. You do not need an ICE ...

Isha_Maniyar_Dis2E

I was confused about part B as well. I figured the I2 (s) would serve as one of the electrodes. For part C, however, HCl is included in the form of H+. In the solution, HCl will immediately dissociate to form H+ and Cl-. Because Cl- is a spectator ion, it will not be included in the oxidation half-r...

Isha_Maniyar_Dis2E

Keep in mind that the species on the left of the double bar (the salt bridge) is not always the anode, and the species on the right of the double bar is not always the cathode. To determine which one is which, you have to determine the flow of the electrons by writing half reactions. Hope this helped!

Isha_Maniyar_Dis2E

It's possible that you have to balance the equation, because you don't seem to have the same number of electrons transferred in both half-reactions. I would suggest writing the half reactions, including the electrons so it is easier to see the flow and how many are transferred.

Hope this helped!

Isha_Maniyar_Dis2E

You can relate Gibbs free energy to K using the following equation:

∆Gº = -RTlnK

As you can see, this is not dependent on pressure or temperature. However, the max work that can be done is given by ∆G at constant pressure and temperature.

Hope this helped!

Isha_Maniyar_Dis2E

The cathode has a negative charge because of the electrons traveling toward it (remember, the electrons travel when delta G is positive, aka a NONspontaneous reaction – work is being done). Because of this, the cations will be attracted to the negative charge of the cathode and travel toward it. Hop...

Isha_Maniyar_Dis2E

We also derived different forms of the Nernst equation and related the equation to K. An important takeaway from class was that when your system is at equilibrium, it is no longer doing work and your battery is dead. So, Ecell will be 0.

Hope this helped!

Isha_Maniyar_Dis2E

Yes. q will always equal 0 for an adiabatic system because of the nature of the conditions. Heat is not allowed to transfer between the system and its surroundings, which is exactly what q measures! So, its value in adiabatic systems will always be zero.

Hope this helped!

Isha_Maniyar_Dis2E

Although I'm not too sure either, my best guess would be to use the deltaG = deltaG(naught) + RTlnQ equation. Solve for each part, relating Ecell to deltaG (I believe we will go over this in week 7 lectures).

Hope this helped!

Isha_Maniyar_Dis2E

When G is negative, the reaction will occur spontaneously. However, when it is positive, energy must be put in for the reaction to occur. When G is 0, the reaction is at equilibrium (this is when your battery would be dead).

Isha_Maniyar_Dis2E

It's true that H+ ions make the solution acidic and more OH- ions make it basic. However, in the balanced reaction, I'm guessing you're confused about including H2O or not. In acidic solutions, we don't really include it – instead, we just show the dissociation. In basic solutions, we write out the ...

Isha_Maniyar_Dis2E

Halogens are always -1, oxygen is (almost) always -2, and hydrogen is always +1. In general, earth metals are +1 and alkaline earth metals are +2.

Hope this helped!

Isha_Maniyar_Dis2E

deltaS is referring to the entropy of the system. deltaS(total) refers to the entropy of the S(system) + S(surroundings).

Remember that S(total) = S(sys) + S(surr).

Hope this helped!

Isha_Maniyar_Dis2E

4A through 4J will be on the midterm; however, there are some sections we do not need to know (e.g. the Born-Haber cycle). Also, we only need to know the basics of Gibbs free energy, as we haven't gone into enough detail on it in class.

Hope this helped!

Isha_Maniyar_Dis2E

There is always external force present, even in a reversible reaction. If there was no external force, there would be nothing to push against, and therefore no work being done. The thing to pay attention to is whether the external pressure is staying CONSTANT. With reversible pathways, the external ...

Isha_Maniyar_Dis2E

If there are more moles of gas on the product side than the reactant side, then the reaction did expansion work (expanded to more moles of gas). On the other hand, if there are fewer moles of gas on the product side, this means the system was compressed. When calculating the number of moles of gas o...

Isha_Maniyar_Dis2E

Entropy is a maximum at equilibrium because deltaS(system) = -deltaS(surroundings). At equilibrium, one side will gain all the possible entropy, while the other will lose it.

Hope this helped!

Isha_Maniyar_Dis2E

Because entropy is a state function, it doesn't matter which pathway we take, because we will end up getting to the same end result. Reversible pathways are more convenient to calculate for, so we tend to use that. In reality, irreversible pathways will have a slightly lower entropy than reversible....

Isha_Maniyar_Dis2E

Aqueous is more entropic than liquid because it is dissociated; however, it still has much less entropy than a gas at the same temperature.

Hope this helped!

Isha_Maniyar_Dis2E

You have to read the conditions of the question to be able to tell if it is open, closed, or isolated. Evaluate what is allowed to be exchanged with the surroundings. Can energy be exchanged, but matter cannot? This would be a closed system.

Hope this helped!

Isha_Maniyar_Dis2E

To add on, you can look at the graph in the textbook to get a better idea of the work being done in reversible v/s irreversible processes. In reversible processes, we are increasing the volume little by little and allowing the external pressure to adjust to equal the internal pressure. As a result, ...

Isha_Maniyar_Dis2E

Residual entropy is the difference between the entropy in a state that is NOT at equilibrium and the entropy of the most stable state of a substance close to absolute zero. I was a bit confused as well.

Isha_Maniyar_Dis2E

We aren't expected to go into this class with knowledge of calculus, so I don't think we will have to actually solve any integrals. Lavelle may, however, ask us to do a problem in a different way that will essentially give us the same answer an integral would – the area under a curve. Hope this help...

Isha_Maniyar_Dis2E

There is an open system, a closed system, and an isolated system. In an open system, energy and matter can exchange between the system and the surroundings. In a closed system, matter cannot exchange, but energy can. In an isolated system, neither can exchange.

Hope this helped!

Isha_Maniyar_Dis2E

The first law of thermodynamics states that energy is never removed or destroyed, it is only transferred. This is helpful because we know that if energy (q) is leaving the system, it MUST be going into the surroundings. We can also use this law to make the statement that + q = - q. This essentially ...

Isha_Maniyar_Dis2E

Along with the tricks mentioned above, you can also determine whether work will be positive or negative based on expansion. If the system is expanding outward, work will be negative (w = -P(delta V)). On the other hand, if it is expanding inward, work will be positive.

Hope this helped!

Isha_Maniyar_Dis2E

For questions with constants such as the 1 atm and 25 degrees Celsius, you do not consider these in determining sig figs. A good tip is to just follow the sig figs in the number given in the problem (molar amount, liters, etc). However, when calculating your answer, keep a lot of decimal places so y...

Isha_Maniyar_Dis2E

The reaction will occur and eventually, we will have a homogenous reaction. However, in this time, heat may be lost to the surroundings. The stirrer allows us to mix the solution and make it homogenous faster so that the temperature is distributed evenly and our measurement may be more accurate. Hop...

Isha_Maniyar_Dis2E

I don't think there is an answer key for the final. Your best option might be to just search up the answers or compare with friends. It might also help to do the problem again and see if you get a different answer!

Hope this helped :)

Isha_Maniyar_Dis2E

You are correct in saying that bonds are breaking. However, you need to take into account the AMOUNT of energy needed to break these bonds. The reverse reaction would be negative because the energy needed to form the bonds in the products (releasing energy, negative G) may be greater than the energy...

Isha_Maniyar_Dis2E

How did you calculate q for the ice? You should have added the q for the ice undergoing a phase change to the q for the ice melting to become 45 degrees. Then, you set this value to mC(delta t) of the 400.0g of water. Solve for the final temperature, given that the initial is 45 degrees Celsius. Hop...

Isha_Maniyar_Dis2E

You know that K decreases as temperature increases. This means that the reaction is exothermic. Because of this, the reactants are favored, and increasing the temperature will result in a shift to the reactant side. More ammonia will NOT be formed, because this is a product. You do not need an ICE t...

Isha_Maniyar_Dis2E

The conversion of bar to atm will be given in the formula handout we are given in each test. It is mainly a difference in values of R. If you are given values in bar, use the value for R involving bar, and then convert to atm using the formula sheet.

Isha_Maniyar_Dis2E

No; Le Chatelier's principle only refers to stressors such as changing pressure or temperature or concentration.

Isha_Maniyar_Dis2E

This is an example of a hydrated complex. We haven't learned this in class, so I wouldn't worry about it. It is a topic that combines coordination complexes being hydrated in water (recall that it typically has 6 binding sites, which is why 6 H2O's can bind).

Hope this helped!

Isha_Maniyar_Dis2E

You can ignore the amount of C(s), because this does not affect the equilibrium (it is a solid). First convert the moles of gaseous H2O to concentration (moles / L). This is your initial. You can then complete the ICE table accordingly.

Hope this helped!

Isha_Maniyar_Dis2E

Practice problems are definitely helpful, but make sure to go over your notes. Also, when you take notes in class, make sure you pay special attention to the things Lavelle mentions but doesn't put up on the slides. He tends to include some questions that require information on these minute details,...

Isha_Maniyar_Dis2E

First find the number of moles of KOH in 2 mL of 0.175M solution. Then, divide by 0.500 L to find the concentration of KOH. The concentration of KOH equals the concentration of OH- in the solution because KOH is a strong base and will dissociate completely into K+ and OH- ions. So, you can find pOH ...

Isha_Maniyar_Dis2E

a) the reactants are favored
b) the reactants are favored (ignore the carbon because it is a solid, so there are fewer moles of gas on the reactant side)
c) the reactants are favored
d) neither are favored (equal number of moles of gas on both sides)
e) reactants are favored

Isha_Maniyar_Dis2E

I was also confused by this, but I'm assuming that you should use the most specific option. For example, if the question is specifically referring to concentration values, it would be best to use K_c rather than K, because that way you know it's specifically for concentration.

Isha_Maniyar_Dis2E

Cl 2 (g) <---> 2Cl (g) K = 1.0 * 10 -5 I 0.001 0 C - x + 2x E 0.001 - x 2x The Cl 2 conc. can be found by doing (2.0 * 10 -3 mol Cl 2 ) / (2.0 L) = 0.001 M Assuming you got this ICE table, you shoukd get the equation: 1.2 * 10 -7 = (4x 2 ) / (0.001 - x) You should then get the quadratic equation fol...

Isha_Maniyar_Dis2E

You have to look at the moles of reactants and products that are in GAS form. There is only one mole of gas on the reactant side because C is a solid. On the other hand, there are two moles of gas on the product side. So, the reactant will proceed to the reactants.

Hope this helped!

Isha_Maniyar_Dis2E

Do we have to use bar as the unit to calculate equilibrium constant? You can use either bar or atm. The units of pressure only matter when you are converting to concentration and you need to use the PV = nRT equation. In that case, you'll have to use the correct R, which has different values based ...

Isha_Maniyar_Dis2E

P has nothing to do with whether we have the values or not, it just indicates that we are talking about partial pressure instead of concentration. Usually, you use partial pressure when the substances in the reaction are gases (as opposed to aqueous solutions).

Hope this helped!

Isha_Maniyar_Dis2E

It's a similar concept as determining the formula for K. For example, if you have Br2Cl2 --> Br2 + Cl2, you would write K as: K = [Br2][Cl2] / [Br2Cl2] Even though you're adding Br2 and Cl2 in the reaction, you multiply them in K. So, adding translates to multiplying when calculating K. Hope this he...

Isha_Maniyar_Dis2E

I assume you're referring to that problem on Q in the textbook. Generally, for Q, you are observing a reaction at a certain point, so you are only looking at a certain direction of the reaction

Hope this helped!

Isha_Maniyar_Dis2E

s-character gives the contribution of the s-orbital to the bond (sp 3 = 25%, sp 2 = 33%, sp = 50%, etc.). As s-character increases, bond angle also increases (lone pairs decrease bond angle and decrease s-character because there are more regions of electron density and the s-orbital does not contrib...

Isha_Maniyar_Dis2E

Higher oxidation states of transition metals result in a higher electron density. The d-block metal pulls the electrons in the bond with the oxygen toward itself, making the bond between the oxygen and the hydrogen weaker. The hydrogen can thus be removed more easily, making it a stronger acid. Hope...

Isha_Maniyar_Dis2E

Yes. A pi bond is specifically made up of unhybridized p-orbitals that overlap side by side. Sigma bonds are formed by hybridized orbitals.

Hope this helped!

Isha_Maniyar_Dis2E

If x is less than 5% of the initial value, you can ignore the " - x" step in the ICE table because the subtraction is negligible. If it is over 5%, you must account for this subtraction.

Hope this helped!

Isha_Maniyar_Dis2E

Only the number of atoms with the lone pair count toward the coordination number. For example, water is monodentate, even though oxygen has 2 lone pairs. The transition metal will bond to the one oxygen atom, not the individual lone pairs.

Hope this helps!

Isha_Maniyar_Dis2E

For starters, you should memorize the strong acids (HBr, Hi, HCl, HClO3, HClO4, H2SO4, HNO3) and the strong bases (LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, Ba(OH)2. This will help you figure out a lot of the Bronsted acids and bases. It also helps to know conjugate acid/base pairs, such as NH4...

Isha_Maniyar_Dis2E

Intermolecular forces are between molecules and they are the forces of attractions between separate molecules that cause them to form liquids and solids. Intramolecular forces are forces which hold atoms together to form a molecule (like a chemical bond). As Michelle said, INTERmolecular = between ...

Isha_Maniyar_Dis2E

For halogen acids (H-X, where X is a halogen), the atomic radius of X will result in a stronger bond between H and X. This will weaken the acid because it won't dissociate as easily in water. As a result, it will be a weaker acid and a stronger base. For oxyacids (H-O-X, where X is a group 5A, 6A, o...

Isha_Maniyar_Dis2E

Some people have the PDF for the seventh edition Chemistry textbook. Maybe someone can link it?

Isha_Maniyar_Dis2E

From what I remember from lecture, monodentate ligands are just ligands that do not have more than one bonding site. This has nothing to do with forming single or double bonds.

In other words, there is nothing stopping a pi bond from being polydentate.

Hope this helped!

Isha_Maniyar_Dis2E

Either can occur; they are both equally likely structures. I think it'll just depend on the question. For example, if the question asks you to draw the Lewis structure such that C 2 H 2 Cl 2 is polar, you would draw the structure so that the Cl atoms are bonded to the carbons on one side and the H a...

Isha_Maniyar_Dis2E

The number refers to the outermost energy level. In the example you showed, the outermost energy level, where all the valence electrons are residing, is n = 2. So, the valence electrons from this ring will interact and form hybrid orbitals.

Hope this helped!

Isha_Maniyar_Dis2E

As you go from left to right, electronegativity will increase (aka it will take require more energy to remove an electron). This is because the valence electrons are experiencing a stronger pull from the nucleus. As you go from top to bottom, electronegativity decreases because the valence electrons...

Isha_Maniyar_Dis2E

For our purposes, all single bonds are hybridized. These are called sigma bonds. This is not a phenomenon; rather, it is an accepted part of chemical structures that in order for covalent molecules to share electrons, they must have hybrid orbitals. For double bonds, one bond will be hybridized, and...

Isha_Maniyar_Dis2E

That's correct. By drawing the Lewis structure, we can determine the VSEPR model (shape/ angle of bond/ general VSEPR formula). We can then determine the hybridization of the central atom by looking at the number of bonding pairs and lone pairs surrounding the central atom. This is why we built up t...

Isha_Maniyar_Dis2E

H 2 SeO 4 has hydrogen bonding, because it has an OH bond. To differentiate between dipole-dipole and dipole-induced dipole, look at polarity. If the molecule is polar, it will do dipole-dipole bonding because the differences in electronegativity will allow attractions between opposite forces. If it...

Isha_Maniyar_Dis2E

First determine the types of intermolecular forces solid Xe and solid Ar have. Neither are polar (considering it's just a single atom), so they cannot do dipole-dipole or hydrogen bonding. This leaves us with London dispersion forces. When determining the strength of London dispersion forces, look a...

Isha_Maniyar_Dis2E

Ionic and covalent bonds are relevant as INTRAmolecular attractions (within the same molecule). However, melting point concerns INTERmolecular attractions (between two molecules). When something melts, its individual molecules remain the same, but the bond between different molecules is weakened. If...

Isha_Maniyar_Dis2E

As boiling point increases, vapor pressure increases. If intermolecular attractions in the liquid are weak, its vapor pressure is high (it will vaporize easily). However, if the intermolecular attractions are strong, it will require more heat to break them, and so the boiling point will be higher. H...

Isha_Maniyar_Dis2E

We will probably eventually have to know the names, as well as the expected bond angles, of VSEPR. Further, we'll have to know the hybridization, or hybrid orbitals, of a given atom in a molecule. But Dr. Lavelle will probably go over all this in a lot of depth before expecting us to know it. Hope t...

Isha_Maniyar_Dis2E

I think we should mainly focus the homework on 3F, because this talks about intermolecular attractions, and we just started this topic in lecture. As we get into VSEPR, we'll be able to use problems from the next unit.

Hope this helped!

Isha_Maniyar_Dis2E

I haven't heard that, but if I had to guess, I would say it's because of the electron-electron repulsion. The atom with the lone pairs will push away from the atom it's connected to because the lone pairs exert a repulsive force. I have heard that the bond angle will be smaller for this reason. For ...

Isha_Maniyar_Dis2E

Does anyone know if we'll be tested on the midterm concepts in the final? Is the final a cumulative of all concepts we've learned throughout the course? If so, I'd like to keep practicing the ideas we covered on the midterm so I don't forget them by the final :/

Thank you!

Isha_Maniyar_Dis2E

Anions that are very polarizable will be large and electron-rich. This means they will have a low charge. The order is O2- < N3- < Cl- < Br- because O2- is electron-rich and the smallest anion given. Br- is the largest anion given (significantly larger than Cl- because of its extra shell), and is th...

Isha_Maniyar_Dis2E

Dr. Lavelle said because there are so many students in the class, it should take about a week (maybe a day or two longer). I'm pretty sure the TA's are grading our papers, so it shouldn't take over two weeks.

Isha_Maniyar_Dis2E

It depends on how long we spend on a unit. If the concept is still relevant in lecture, you can do questions on it and turn it in for credit.

Hope this helped!

Isha_Maniyar_Dis2E

A lot of high school chemistry classes taught VSEPR; however, don't worry if you don't know them. I'm sure Dr. Lavelle is going to go over it with us. The table is in the textbook I think.

Hope this helped!

Isha_Maniyar_Dis2E

The hydrogen bond is very strong (stronger than a dipole-dipole interaction) because of the huge difference in electronegativity. When ionic compounds are dissolved in water, the positive dipole end of the water molecule (H+) attracts the anion from the compound, while the negative dipole end of ano...

Isha_Maniyar_Dis2E

P, S, and Cl can accommodate more than 8 valence electrons. This is because atoms in period 3 or higher have d-orbitals in the valence shell that can accommodate additional electrons. These are all the octet exceptions, to my knowledge.

Hope this helped!

Isha_Maniyar_Dis2E

This is more of a concern for tests – do we have to show work when calculating valence electrons? Will Dr. Lavelle care if we just write down the number of valence electrons when trying to determine the Lewis structure of a molecule?

Thanks!

Isha_Maniyar_Dis2E

There are no hydrogen bonds in CHF3. For there to be a hydrogen bond, the F and H must actually be connected to each other. However, the Lewis structure of CHF3 shows C as the central atom, with the three F's and the one H all connected to the C. So, we only look at London dispersion forces in the t...

Isha_Maniyar_Dis2E

A molecule can do dipole-dipole bonding if it is polar. It can do hydrogen bonding if it is polar and if it has an OH- or NH- bond. H-bonding is basically a type of dipole-dipole bonding, except with OH and NH bonds. All molecules have London dispersion forces, which are momentary attractive charges...

Isha_Maniyar_Dis2E

No, we have not covered this equation in class. If we do go over it, I expect it will be later in the course. However, considering it's related to energy, we might just not cover it, as we're already finished with the quantum unit.

Isha_Maniyar_Dis2E

It depends on what orbital the electrons are in. For example, carbon has 4 valence electrons, but they are all in their own orbitals, so none of them are paired. However, if two electrons are in the same orbital and not bonded to another atom, then these are lone pairs. Bonding pairs would be the &q...

Isha_Maniyar_Dis2E

I remember in AP Chem, we learned that P, S, Cl, Br, I, and Xe all had expanded valence shells. This is because the outer shells are very large (atomic radius is large), and can accommodate more than 8 electrons. However, Dr. Lavelle only mentioned P, S, and Cl as having expanded valence shells. Is ...

Isha_Maniyar_Dis2E

When an electron is in its ground-state, this means it is in its lowest-possible energy level. When the electron is excited, it absorbs energy and jumps to an energy level higher than the ground state energy level. Usually, the electron is not stable when it is in an excited state, and will eventual...

Isha_Maniyar_Dis2E

Resonance structures have delocalized electrons, which means the negative charge is spread throughout all the bonds in the molecule. This affects polarity, as the spread charges reduce the polarity of the overall molecule.

Hope this helped!

Isha_Maniyar_Dis2E

a) Pb --> The valence shell is 6p, and there are 2 unpaired electrons in this shell. b) Ir --> there are 2 unpaired electrons (5 orbitals in valence shell, 7 electrons in the shell) c) Y --> 1 unpaired electron in valence shell (4d) d) Cd --> no unpaired electrons (5 orbitals in valence shell 4d, 10...

Isha_Maniyar_Dis2E

Adding on to all the previous comments, the X, Y, and Z in the electron configuration indicate the different planes. I also think Dr. Lavelle said it was a good idea to specify this because it shows that the electrons are unpaired. For example, when you write 1s2 2s2 3px1 3py1 3pz1, you are showing ...

Isha_Maniyar_Dis2E

For tantalum, the entire 4f orbital is completely filled (14 electrons). We know this because the f-block falls in the periodic table before we reach tantalum. The 6s orbital is completely filled as well; this just follows the same pattern we've been using for d-block elements (besides the exception...

Isha_Maniyar_Dis2E

I remember my TA saying something about knowing the general range for visible light and UV radiation. Even this doesn't need to be too specific. 400-700nm is the visible spectrum, which might be a good thing to memorize so you can gauge the general range for the other wavelengths. Overall, I wouldn'...

Isha_Maniyar_Dis2E

The quantum numbers all describe a single electron in any atom. So it doesn't have to be an atom with ONLY one electron; Lavelle was just saying that the four numbers all describe a single electron – the energy, shape, orientation, and spin state.

Hope this helped!

Isha_Maniyar_Dis2E

It can also be used to find photons per mole! This conversion is probably good to know, especially for the quantum mechanics unit :)

1 mole = 6.02 * 10^23 photons

Isha_Maniyar_Dis2E

I don't think we will be going into finding the spin state of a specific electron. The idea of that quantum number is just that an electron can either have a positive spin or a negative spin, or a quantum number of +1/2 or -1/2. I think the main concept to understand is that if given 2 electrons in ...

Isha_Maniyar_Dis2E

Based on AP Chemistry, we rarely had to convert to units of Angstrom. That being said, Angstroms are most used to describe bond length, which I believe is our next unit. We may be expected to use it most in that unit.

Hope this helped!

Isha_Maniyar_Dis2E

I think a "quantum" is an amount, whereas a photon is an actual bundle of energy. This is why a photon is also known as a "quantum of energy".

Hope this helped!

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