CHEMISTRY COMMUNITY

Justin Seok 2A

Math wise, I do not believe that a pseudo first-order reaction differs from a regular one, but I believe, like with water, having a concentrated solution means that there is a much higher concentration of C12H22O11 in a given solution than H2O, so your rate law only takes into account H2O, since tha...

Justin Seok 2A

For standard Gibbs Energy, you just use -nFE. And for knowing which one is the cathode and anode, you have to know the standard E values for both so you can determine which one is lower and thus the anode.

Justin Seok 2A

Yes, the anode is always the one with the lower concentration.

Justin Seok 2A

Intermediates are a necessary part of the reaction, as the reactants don't necessarily just form into the products in one step so intermediates are formed and used up within reactions. However, they do not show up in the products.

Justin Seok 2A

Expansion usually causes an increase in work, as the volume is increasing against an opposing pressure.

Justin Seok 2A

Yes, because C is zero order, its concentration doesn't matter. So when comparing experiments 1 & 4, where the only concentration changing is that of C, because the rate does not change you can presume that change in [C] has no effect on the rate so it is not involved in the rate law.

Justin Seok 2A

I believe it may be the way that ln works with negatives, since -ln([A]t/[A]0) would be equivalent to ln([A]0/[A]t). So the equation would go from [A]t=[A]0*(e^(-kt)) to [A]t/[A]0 = e^-kt to ln([A]t/[A]0) = -kt to t = ln([A]0/[A]t)/k.

Justin Seok 2A

Since the first step is pointed out to be the slowest, the rate law for that reaction determines the rate law for the entire reaction. Thus rate = k[NO][BR2] would be the rate law, not the rate law for the whole reaction.

Justin Seok 2A

Qualitatively endothermic reactions increase K when temperature increases while exothermic reactions lower K when temperatures increase, and the opposite when temperatures decrease. You can use the Van't Hoff equation to figure out how much exactly K values change when temperatures change.

Justin Seok 2A

By looking at the redox equations, you can find n by looking at the amount of electrons transferred, like Cu ---> Cu2+ + 2e-. In this case n=2 since 2 electrons are being transferred.

Justin Seok 2A

For example in a redox eq like Cu2+ + 2e- ---> Cu, n would be 2 since 2 electrons sre being transferred.

Justin Seok 2A

You can add OH- after adding your protons so that the OH- and H+ cancel out in order to form H20.

Justin Seok 2A

It would probably be oxidation as the half-reaction would look like 3I- --> I3- + 2e-, so electrons are lost. However, I am not completely sure if this is how this reaction would work.

Justin Seok 2A

You can figure that the Cl would have a charge of -1 because it is in the 2nd last row of the periodic table. So Cl2 would have a total charge of -2, meaning that Hg2 would have to have a charge of +2 in order to keep Hg2Cl2 neutral and Hg would have a charge of +1.

Justin Seok 2A

H+ seems to be more commonly used and doesn't involve oxygen, which may complicate things balancing wise.

Justin Seok 2A

Essentially, this problem is asking for which equation would cause the least entropy so that G and H are similar. Entropy changes quite significantly when phases change, so the equation with the least entropy change would be the one with the solids on both sides.

Justin Seok 2A

I believe that boltzmann's constant is already in J/K, so I would assume that because the equation involves ln W, the units of W are not involved in calculations, thus leaving the units of boltzmann's constant, J/K, the same for the product.

Justin Seok 2A

You would probably need to know either one of the initial or final mole values, and figure out what changes out of pressure, volume and temperature to figure out how your mole amount has changed.

Justin Seok 2A

Yes, O3/ozone has an oxidation number of zero, its formal charge is zero as well.

Justin Seok 2A

I believe so because one of the Cls are involved in the HClO reaction, which increases the oxidation number from 0 to +1, and the other Cl is involved in the Cl- reaction, which decreases the oxidation number from 0 to -1.

Justin Seok 2A

It shouldn't matter where they are placed in the equation, but I suppose that the solutions manual put them at the end simply because they are not as important as the reactants and products in the reducing and oxidizing half-reaction if that makes sense. I would highly doubt that the test graders wo...

Justin Seok 2A

The lecture that day was simply on past exam questions, going over questions on topics like equilibrium, acids and bases, and thermodynamics/chem.

Justin Seok 2A

A reaction that is spontaneous will have a negative Gibbs free energy, while a reaction that has a Gibbs free energy is in equilibrium.

Justin Seok 2A

If you reverse the reaction, then Kc becomes 1/Kc. If you multiply the reaction by a constant, Kc equals Kc^x, where x is that constant.

Justin Seok 2A

An adiabatic process transfers no heat, and a diabatic process transfers heat

Justin Seok 2A

When work is done on the system, it is positive, and when the system does work on its surroundings, it is negative.

Justin Seok 2A

They are most likely given in one of the tables in Chapter 4. However, on tests/midterm those values should be given if needed to solve a problem. So if a phase change is involved in figuring out the enthalpy of fusion and vaporization those values represent the amount of enthalpy required to change...

Justin Seok 2A

Solids and liquids have less energy, so they are more tightly packed and thus take up less space than gases, which have more energy and are liable to move around more.

Justin Seok 2A

Lyndon posted a Chem Community page titled Pizza Rolls, which should have a helpful practice worksheet up sometime in the evening today.

Justin Seok 2A

The slope of the heating curve is the same as the heat capacity, so if the curve is steep, then the heat capacity is higher, while if the curve is flatter, the heat capacity is lower.

Justin Seok 2A

Reversible processes can be thought of as the maximum work possible done by the gas, as external pressure constantly changes in order to match the internal pressure. Irreversible processes simply have constant external pressure, so the work done would not be as high.

Justin Seok 2A

I think in this case percent ionization would simply be 2.344 * 10^-3/(0.010), since it is usually concentration of acid ionized/initial acid concentration or in this case [F-]/[HF].

Justin Seok 2A

I would assume chemical equilibrium, acids and bases, thermochemistry and possibly some thermodynamics as well, based on how far we get in lecture before the midterm. Based on last quarter in Chem 14A, there are likely to be plenty of review sessions for different topics and a big review session bas...

Justin Seok 2A

The steam and water can have the same temperature but the steam would carry more energy in the form of heat since it takes energy to drive a phase change from liquid to gas. Assuming that a burn would be stronger based on the energy it transfers onto skin, the steam would have a stronger burn.

Justin Seok 2A

Essentially, when a liquid turns to gas, the bonds between atoms must be broken in order to become gas, which would presumably take a lot more heat than turning solids to liquid, which don't completely break the bonds but make them weaker.

Justin Seok 2A

In the book, it says that w = -P∆V when the external pressure is constant, so presumably if the irreversible process involves constant external pressure then yes. It wouldn't be used for reversible processes, as those use w = -nRTln(V2/V1).

Justin Seok 2A

So the simplest way would be to find Pkb using the equation 14=pka+pkb. Once you find your pkb, you can use it to find kb using 10^-pkb.

Justin Seok 2A

Essentially steam carries more energy as it takes a lot of energy to transition from liquid to vapor. So while they may have the same temperature, steam will very likely have a higher heat content due to the heat of vaporization as you mention.

Justin Seok 2A

The Ka is basically the measure of how much something dissociates, so if you have a larger Ka then you know more H3O+ will form, giving you a stronger acid. On the other hand, a lower pKa is a sign of a stronger acid.

Justin Seok 2A

For the most part, in this section, we mostly work with weak acids and such since they are best suited for equilibrium type problems. However, it would probably be good to have a general idea of common strong acids just so you won't get blindsided if they come up on the test. However, it shouldn't b...

Justin Seok 2A

So you can solve for hydroxide/hydronium ion concentration using the other concentration by using the fact that 1.0*10^-14 = [H30+][OH-]. Because you know that the equilibrium constant for water, Kw, is 1.0*10^-14, you can thus solve for either concentration by dividing Kw by the ion concentration t...

Justin Seok 2A

A thermodynamically stable thing tends to keep its form well. So when comparing Cl2 and F2, we simply look to see which of the two has a smaller equilibrium constant. Since Cl2 does, it is more thermodynamically stable, since a smaller ratio of Cl2 dissociates into Cl atoms compared to F2 and F atoms.

Justin Seok 2A

The breaking of a bond requires energy to do, so increasing the temperature, and as a result increasing the energy, of the reaction would cause more X to form.

Justin Seok 2A

I got something similar to you as well when doing 5.35, I think its simply due to the fact that in the solutions manual they use 18 Kpa instead of 17.5 Kpa, since your answer is so close to the answer they give.

Justin Seok 2A

K is pretty much a more ambiguous term than either Kc or Kp. Often times when K is used, it should be pretty apparent whether Kc or Kp should be used, like with Kp when you are given partial pressures or Kc when there are no gas-phase reactants or products.

Justin Seok 2A

Most times you can tell what direction a reaction will go based on the initial concentrations you are given. If you are only given that you start off with reactants, then the reaction will surely go to the right. If given only products initially, then the reaction will go to the left. If you start w...

Justin Seok 2A

The activity level for a solvent is equal to or very close to being equal to 1, so it doesn't affect the equilibrium constant.

Justin Seok 2A

Since change is often unknown in problems using the ICE box, we usually attribute a variable to the amount of change in products and reactants, oftentimes x. Using stoichiometric analysis, like in an example equation 2 A + 3 B <--> C, the change would be -2x for A, -3x for B, and +x for C. Then use ...

Justin Seok 2A

I think that in this problem it is assumed that we are working with partial pressures as the problem is looking for the partial pressures of the gases in the mixtures. However, in other problems where it doesn't say whether or not you are specifically looking for pressure/concentration, it would be ...

Justin Seok 2A

C would have different values for PO2/PO3 for the two samples since only the equilibrium constant, which would be PO2^3/PO3^2, would have the same value between the two samples as it is a constant. It's kind of like a trick question to see if you can recognize that PO2/PO3 is not the equilibrium con...

Justin Seok 2A

It would be true as the equilibrium concentrations would be the same regardless of whether the reaction starts with pure reactants/products since the ratio between the two concentrations is equal to the equilibrium constant, which doesn't change.

Justin Seok 2A

What is the biological importance of the heme complex? The importance of the heme complex, which is a complex of Fe and a tetradentate porphyrin ligand, is that it is part of the protein myoglobin and is responsible for binding to O2 through its Fe cation, since it has one available space to after ...

Justin Seok 2A

Pretty much for all of the parts, there is an acid that reacts with water to release its H+ proton and thus form the conjugate base and a hydronium ion H3O+.

Justin Seok 2A

Oxygen is not necessarily gonna be sp2 or sp3 both times even though they are in the same molecule. Hybridization pretty much equates to how many zones of lone pairs/bonds are on the atom. So the top oxygen is sp2 since it has 1 bond and 2 lone pairs while the other oxygen is sp3 since it has 2 bond...

Justin Seok 2A

cisplatin is specifically used for chemotherapy and stopping cell division due to its cis orientation. Essentially, because cisplatin has both of its Cl's on one side, the two Cl's can bond to a close together guanine pair. It is bidentate since the two Cl's are on the same side, so cisplatin can st...

Justin Seok 2A

A strong acid is one that pretty much completely disassociates into a proton and anion in solution. One example would be HCl, which just becomes H+ and Cl- in solution.

Justin Seok 2A

Since you are given the concentration of H+, it is pretty simple to calculate pH. The equation for pH is just -log[H+], so in this case it would be -log[10^-5] which is equal to 5.

Justin Seok 2A

pOH would be like pH, in that its a measure of concentration. So in problems where a pOH or a concentration of OH- molecules is given, it is likely you will have to use the equation to find pOH to solve for that problem. To find pOH, use the equation pOH = -log[OH-], using the concentration of OH- i...

Justin Seok 2A

The Lewis definition states that an acid is an electron acceptor while a base is an electron donor. The Bronsted definition states that an acid is a proton donor, while a base is a proton acceptor.

Justin Seok 2A

A lot of the strong bases are alkaline earth metal or alkali metal hydroxides, examples including KOH, NaOH, CsOH, (Ca(OH)2)), and so on. This may be why it seems like most bases have OH or O in them, but it is not necessarily a requirement. It does seem to be a trend among strong bases, however.

Justin Seok 2A

Since HCl is a strong acid, the ions that make it up disassociate completely in water, so the image showing individual H+ and Cl- ions would be the most accurate. A weak acid, in contrast, would mostly show the acid in question floating around, with a few disassociated ions floating around, as weak ...

Justin Seok 2A

Essentially, amphoteric compounds can act as both an acid or a base. Water is a classic amphoteric compound as it can donate one of its H+ to become OH- or accept H+ to become H3O+.

Justin Seok 2A

"Coordination compounds are molecules that poses one or multiple metal centers that is bound to ligands (atoms, ions, or molecules that donate electrons to the metal). These complexes can be neutral or charged. When the complex is charged, it is stabilized by neighboring counter-ions. A complex...

Justin Seok 2A

The most recent concept in the lecture that we will have to study for test 2 are the sigma and pi bonds.

Justin Seok 2A

I believe that the first lewis structure would work best from a formal charge perspective, as I generally only has a single bond to keep its formal charge as zero. The I in the middle would have a formal charge of -1 having two single bonds, which would be appropriate as I3 has a negative charge.

Justin Seok 2A

If it were carbon dioxide, the molecule would be nonpolar because the molecule would be linear, thus meaning the dipole moments of the two oxygens would be canceled. However, the carbon monoxide would still be polar since the less electronegative carbon has a more positive charge while the more elec...

Justin Seok 2A

Essentially, a trans molecule will have the same atoms in the bonded molecules on opposite sides, while a cis molecule will have the same atoms in the bonded molecules on the same sides. Thus, a trans molecule will have the dipole moments of the atoms cancelled out, while a cis molecule will not hav...

Justin Seok 2A

No, the l would equal 1 for the 4pz state, since p corresponds to l=1. Generally, s is l=0, p is l=1, d is l=2, and f is l=3.

Justin Seok 2A

I would assume that the explanation would probably have to include the effect of the electron to really get at the heart of why the double bond is shorter than a single bond. A good way to explain it would be that since double bonds have more bonding electrons involved than single bonds, there is a ...

Justin Seok 2A

From my understanding, polarity occurs when the electrons of a molecule aren't distributed equally. An easy way to figure out whether a molecule is polar is to see if its shape is non-symmetric, so H20 would be polar since it is a bent molecule, while BeCl2 is nonpolar since it is a linear molecule.

Justin Seok 2A

Electronegativity increases across a period. Electronegativity is a measure of an element's ability to attract electrons, so those to the right of the period are able to attract electrons more easily as they need electrons to fill their valence shells and thus are the most electronegative.

Justin Seok 2A

Polarizing power also becomes stronger the smaller an atom is since the effective nuclear charge has a stronger effect, while polarizability becomes higher when an atom is larger since its electrons are more easily affected due to a larger distance from the nucleus.

Justin Seok 2A

So Heisenberg's uncertainty equation is basically (uncertainty in position)*(uncertainty in momentum)=h/4pi. Since we have uncertainty in speed, +- 1 m/s, we can find uncertainty in momentum by multiplying it by the mass of an electron, 9.11*10^-31. Then we can divide that value from both sides to g...

Justin Seok 2A

Rb has the larger Atomic Radius, Consequently Sr has a higher polarizing power. Larger radius=more polarizability smaller radius=more Polarizing Power... how does Rb have the larger atomic radius? A good way of looking at the atomic radius is that when going further down a period, elements have mor...

Justin Seok 2A

I don't believe you can find the frequency directly using de Broglie's, at least not using c=wavelength*frequency. However, it may be an intermediate step in a photoelectric effect problem, where you are given wavelength of an ejected photon, which you use to find the velocity of the photon, and the...

Justin Seok 2A

We may be given electronegativity numbers for some of the elements involved in the problem if any do arise in the midterm, however, it would probably be in your best interest to generally know the trends for electronegativity, and other periodic trends as well. In the case of electronegativity, it g...

Justin Seok 2A

Essentially, you can tell whether an electron/photon will be emitted or absorbed based on whether n goes up or down. E for an electron is measured negatively, meaning that when n approaches infinity, E approaches 0. When n=1, the lowest quanta it can be, E becomes -hR/n^2, the most negative value E ...

Justin Seok 2A

In the answer key it doesn't show any resonance structures for c and d so I guess they are not needed for the problem.

Justin Seok 2A

From what I understand, resonance structures are less how the bonds are really like in real life but more the best representation of the idea of delocalized electrons within a bond/molecule. So instead of having different bond lengths between double and single bonds within the resonance structures, ...

Justin Seok 2A

For the p orbitals, the nodes exist on the x-plane, y-plane, and z-plane and for the d orbitals, the nodes exist on the xy, xz, yz, x^2-z^2, and z^2 planes, and occur when the wave function goes from positive to negative or vice versa, as that is where the function, the probability of electron densi...

Justin Seok 2A

I believe that generally the same topics will be reviewed in the same topic review sections, i.e. atomic spectra, photoelectric effect, etc. in Early Quantum Review sections, but the lecturing styles/examples given by the TAs may be different depending on who is leading the sessions. So if you are e...

Justin Seok 2A

I believe that at least some of the conventions for nomenclature are contained in Fundamental section D. I would assume that we wouldn't need to know it that well or that it will be given since we weren't assigned this reading, but may still be helpful to know for the future.

Justin Seok 2A

It may be useful to generally know the order of common waves, like the fact that ultraviolet is lower wavelength than visible light for things like Lyman series or infrared is higher wavelength than visible light. It probably won't be super important to remember though.

Justin Seok 2A

p represents the specific subshell l=1, the same way that s is l=0, d is l=2, or f=3. So if you are given 6p, you know that the l is 1 due to the presence of the p.

Justin Seok 2A

I believe at n=5 is the minimum value of n that has a g-subshell.

Justin Seok 2A

You are correct, the possible values for l should be 0-5, as n=6 shouldn't be an exception. Maybe you were reading the answers for another problem as this one is 1D.14, which shouldn't have an answer in the solutions manual.

Justin Seok 2A

Also for orbitals in general, n is the energy level, l is the subshell and ml is the orbitals within the subshell. So if you are only given n=2 for example, you would know that l can either be 0 or 1, or s or p, since l is 0, ... , n-1. For l, 0/s has 1 orbital, 1/p has 3 orbitals, 2/d has 5, and 3/...

Justin Seok 2A

In our discussion, I believe the TA said that the most important thing to note would be that visible light occurs at 400-700 nm. Personally I think it would be good to know that ultraviolet occurs below 400 nm and infrared occurs above 700nm for random things like Lyman series, and it might be helpf...

Justin Seok 2A

Just for more info, l=0 is s, l=1 is p, l=2 is d, and l=3 is f. So when you have a certain energy level, let's say n=2, then the corresponding l values are 0 and 1, the values that are less than or equal to n-1. s has 1 corresponding orbital, p has 3, d has 5, and f has 7, and these corresponding or...

Justin Seok 2A

Based on the information given, the 102.6 nm, we know that emission is ultraviolet and thus initial energy level=1, indicative of the Lyman series. Given that we have a wavelength, we can find the frequency using c=frequency*wavelength and it turns out to be 2.922*10^15 s-1. Frequency = R(1/n1^2 - 1...

Justin Seok 2A

Pretty much what you said, to find wavelength, you can simply use the value of frequency and the equation c=(wavelength)(frequency) to find wavelength.

Justin Seok 2A

The units for joules are kg*m^2/s^2. It is the unit for Energy, which is pretty much the force used times meters moved(I think), which would be mass times acceleration(m/s^2) times meters, leading to the units for joules. In chem, it would be used as a unit of energy for an electron or photon, for e...

Justin Seok 2A

I think that the "electric field" would be referring to the electric field waves that are oscillating. So c would be correct because when the frequency lowers, the field/waves would be "changing" more slowly at a given point since the waves simply don't oscillate as much per seco...

Justin Seok 2A

I believe my TA told our discussion group that they would be providing like a printed exam and you would be writing directly onto that to answer, so probably no blue book/scantron needed.

Justin Seok 2A

Not 100% sure either but I think it means that even though the kinetic energy is close to zero, the electron would still sort of be drift to the detector as the detector has a positive charge and the electron is negatively charged, thus attracting it towards the detector based on charge. I could be ...

Justin Seok 2A

I also wasn't 100% clear on this question, but I kinda took it to mean that since there aren't as many oscillations, the electrical field wouldn't change as quickly since it takes a longer time for each wave to oscillate. Not the most technical explanation but hopefully it kinda helps clear things up.

Justin Seok 2A

Technically, increasing the intensity would actually increase total energy, as there are simply more electrons being emitted. However, what intensity doesn't change is the amount of energy per photon/electron being emitted, as that depends on frequency only and is based on the equation E = hv.

Justin Seok 2A

I would assume that the equation you would use in this situation would be 1/2mev^2 = Energy - workfunction. Given that you know both the velocity and the work function already, you can solve for energy needed by adding 1/2mev^2, where me is the mass of one electron, and the work function.

Justin Seok 2A

The number of molecules in the solution after all of the dilutions would be the same, however, the question asks for the number of molecules in 10 ml of soln, which would be now a small percentage of the total volume due to the ninety dilutions. So in order to solve for this, what I did was find the...

Justin Seok 2A

I think that the word "combustion" in the problem allows you to infer the presence of oxygen gas in the equation. And with carbon monoxide, it is formed during incomplete combustion when there is not enough oxygen to form carbon dioxide, but I would guess that for simpler problems where in...

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