CHEMISTRY COMMUNITY

Ronald Yang 2F

I'm pretty sure it's referring to the substituents. They are drawn in a way that makes them "trans." For cycloalkenes, they are technically "cis" since the priority would be given to the ring.

Ronald Yang 2F

It's because two intermediates exist at that "intermediate" phase. Together, they make up that amount of energy considered "intermediates" on the reaction profile.

Ronald Yang 2F

By definition, an aldehyde is R-(C=O)(H), with the H attached to the carbon. Thus, the double bond is implied. You would only do "enal" if there were another double bond in the parent chain.

Ronald Yang 2F

Oh, I see. Yes, that would be correct too. I guess it depends where the substituents are placed. What I said would apply to a molecule like 1,3-dimethylcyclohexane, while yours would apply to a molecule like 1,4-dimethylcyclohexane.

Ronald Yang 2F

Not necessarily. If the cis molecule has its substituents in equitorial positions, it would be more stable than a trans molecule with a substituent in an axial position and a substituent in an equitorial position.

Ronald Yang 2F

It wouldn't matter as long as you determine whether it's axial or equatorial. Let's say you draw methylcyclohexane and put the methyl group axial up on "Carbon 1," whichever carbon you designate the first. If you were to view the molecule at another angle, you could rotate your view 180 de...

Ronald Yang 2F

(CH3CH2)2NCH2CH2CH3 The thing is, the way you're doing it is that you're taking one of the (CH3CH2) groups and taking that CH2 and putting it on the other (CH3CH2) group, leaving the "methyl" group on the N that you are referring to. It groups the (CH3CH2) for a reason because they stay to...

Ronald Yang 2F

We don't count "tert" in the naming for alphabetical, so, because the substituents are in "equal" positions, we put tert-butyl with the lower number then methyl, since b>m alphabetically.

Ronald Yang 2F

It's wrong. It's supposed to have 1-bromo-3-iodo. Priority only applies to alkenes in determining cis/trans, not for naming for numbering.

Ronald Yang 2F

If you look at each carbon on each side of that double bond, there is a different amount of carbons connected to each one (left of double bond is connected to three carbons, while right of double bond is connected to two carbons). We know that that double bond will act as a nucleophile to attack som...

Ronald Yang 2F

Hey Derek,

I made a mistake. The priority I was talking about only applies to alkenes in determining "priority" for whether a molecule is cis or trans. So sorry about that. You are correct, and the workbook has it wrong.

Ronald Yang 2F

Yes, both are correct. Since "di" is not counted for the ordering, "methyl" would come before "methylethyl."

Ronald Yang 2F

I think the bigger the functional group and the greater the atomic number of the element in that functional group connecting with the rest of the molecule (usually shown as an "R" group) is a good indication of priority. It's also best to specify that one for the "-ol" because th...

Ronald Yang 2F

There are many ways you can draw C8H16. Cyclopentane with an isopropyl substituent is one form, while cyclooctane is another. Like she did in lecture, when given a formula, there could possibly be more than one way to draw the structure. Both are correct ways to exemplify C8H16. With regard to if on...

Ronald Yang 2F

The orientation wouldn't matter in this case, because whether you start counting from the right or left of that condensed formula doesn't change the numbering specified in the formula. So, both are correct.

Ronald Yang 2F

1) It's fine to do it like that. Actually, adding the double bond shows your understanding that it's an alkene.

2) It's not incorrect. It's essentially the same thing. As long as the structure is the same, doesn't matter if its flipped.

Ronald Yang 2F

Actually, cis or trans doesn't apply for this molecule because it's an alkane, not an alkene. The structure you drew right next to the answer is correct as well. It's just that the left one shows better the "tetrahedral" shape of that sp3 carbon.

Ronald Yang 2F

Usually, the frequency factor A would be considered equal to 1 if not specified, making ln(A)=ln(1)=0, though this may not always be the case since it depends on the specific orientation of a collision. We usually will use the Arrhenius equation to compare two rate constants at their respective two ...

Ronald Yang 2F

"Butanol" refers to the parent chain; chlorine is just a functional group, so you wouldn't use the alphabetical rule. Now, if you were referring to the priority of functional groups, then the OH would have priority over the chlorine, which is why the numbering is how it is.

Ronald Yang 2F

First of all, not sure how you determined that bromine was the "first substituent," but, if you refer to the Ochem textbook, for functional groups like halogens (Cl, Br, I), priority is based on the atomic number of the atom, so the greater the atomic mass, the greater is its priority. Thu...

Ronald Yang 2F

Propane

Ronald Yang 2F

The (Z) is essentially like (cis) for a molecule where we could not determine if it was cis or trans due to the fact that there are more than two groups off the double bond, so we use E or Z instead (respectively similar to trans or cis). Trick: Z, Zame, Zide (same side :D). If you were unsure about...

Ronald Yang 2F

I don't think the order of the reaction for this problem matters because you are comparing two different situations of the same reaction with the same order. When you use the Arrhenius equation to find the rate constant, to find the factor, you would divide the "new" rate law with the &quo...

Ronald Yang 2F

Because the amount that the concentration of B increases is essentially the amount that the concentration of A decreases (since you use up A to make B). Thus, you subtract the value you find with the mole ratio from the original amount of A to figure out the amount of A after that 115 seconds pass.

Ronald Yang 2F

Also, if you look at a reaction mechanism and a species that appears in the reactants of one elementary step appears again in the products of a following elementary step, then most likely that species will be a catalyst, as it is reformed.

Ronald Yang 2F

When the double bond "attacks" the partially positive H, essentially the electrons from the H-Br bond will go to the Br, meaning that when that H makes a bond with one of the carbon atoms, in this case being the carbon on the end because we want to put the positive charge on the carbon tha...

Ronald Yang 2F

Technically yes, since depending on which carbon you start from determines how long it would be. For example, you would want to start on a carbon that is preferably at the "end" of a chain, meaning that you would start counting from a primary carbon (carbon attached to only one carbon).

Ronald Yang 2F

The fact that it says "Nu" identifies it as a nucleophile. That curved arrow just refers to a transfer of electrons from ":Nu" to the electrophile, which is probably the carbon cation. Also, nucleophile and electrophile would be used for specific atoms involved in the electron tr...

Ronald Yang 2F

For elementary steps in a reaction mechanism, we just look at the molecularity to determine the overall order. So, basically, the number of reactants of a certain species determines the order of that reactant. Thus, it would be first order for O3 in step 1, and it would be first order for O3 and fir...

Ronald Yang 2F

The main thing is that the metal electrode itself represents a solid species, and cell potential only changes when a reaction is deviant from equilibrium. Since Q depends on only aqueous species, increasing the concentration of the solid species won't change Q, meaning that the reaction would remain...

Ronald Yang 2F

When change in T is zero, that doesn't mean q=o. When temperature is constant, delta U is zero, shown by the relationship delta U = 3/2*n*R*delta T. An application to this is for a reversible, isothermal expansion. Since delta U=0, q+w=0, so q=-w. In this case, q is not equal to zero. In fact, heat ...

Ronald Yang 2F

A cell that has an oxidation half-reaction and a reduction half-reaction, both of which have equal to opposite voltages (ex. a concentration cell).

Ronald Yang 2F

Because the Cl- is necessary for the oxidation half-reaction to occur. You also include species that is needed in the half-reaction. Also, on the right side resembling the cathode, isn't it Hg₂²⁺ -> Hg?

Ronald Yang 2F

I'm guessing you are talking about one compound being both reduced and oxidized into two products. In a sense, it is also the reducing agent and oxidizing agent. For example, consider the reaction 2H 2 O -> 2H 2 + O 2 . Water is reduced to hydrogen, while water is oxidized to oxygen. If it's easier ...

Ronald Yang 2F

No, you are correct. You must flip the half-reaction resembling reduction that would occur at the anode to yield the half-reaction resembling oxidation. The reason why they leave the "oxidation" half-reaction in its reduction form is because they want to emphasize that the standard cell po...

Ronald Yang 2F

Usually, when we have MnO4-, it is reduced to Mn2+. That's just a rule of thumb. Potassium is a spectator ion, so I believe that's why we don't take that into account. Similarly, because chlorine is neither oxidized or reduced in this half-reaction, it just acts as a spectator ion, not necessary for...

Ronald Yang 2F

Well, it's asking for the molar entropy, so number of moles does not need to be specified.

Ronald Yang 2F

I don't think so because then the system would transfer energy to the surroundings in the form of work - wouldn't be isolated then.

Ronald Yang 2F

Like the explanation says, "It is not immediately obvious." It is after you calculate the entropy change of the reaction and yield a negative value. This means that the sum of the absolute entropies of products minus the sum of the absolute entropies of reactants equals a negative value, m...

Ronald Yang 2F

delta G = delta H - T times delta S. Since the gas is "heated to 350 K" and "cooled back to 298 K," delta H is zero. Since delta H and delta S are both zero, delta G must also be zero.

Ronald Yang 2F

The thing is, the molar heat capacities for water as a liquid and as a gas are different, so the cooling term and the heating term wouldn't cancel out. That's why I think for this kind of situation where you perform vaporization at a lower temperature than that of the boiling point that the total ch...

Ronald Yang 2F

Okay, so ΔS= \frac{q(rev)}{T} . The type of expansion that occurs is isothermal, reversible expansion. Thus, since it's isothermal, ΔU=0, since ΔU=(3/2)*n*R*ΔT and ΔT=0, making ΔU=q+w=0 and q=-w. The work for an isothermal, reversible expansion is -nRTln(V2/V1), and since q(rev)=-w, q(rev)=-...

Ronald Yang 2F

I think it's because, the more complex a molecule is, the higher its molar entropy is. Since the molar mass of benzene is higher than the molar masses of each of the rest of the molecules, it holds the most complexity, and thus, the highest molar entropy. You can also think about it in terms of...

Ronald Yang 2F

By convention, we make work "negative" PΔV because we are looking at the work done by a system. If it were work done on a system, then it would just be PΔV without the negative sign. This is because, from the standpoint of the system, when it expands, hence when ΔV is positive, the system ...

Ronald Yang 2F

I'm looking at question 9.13, and it seems like the gas is nitrogen gas. Nitrogen gas isn't a monoatomic gas. It's a linear molecule, since it's N₂. I'm guessing that's why we don't use the heat capacities for a monoatomic ideal gas.

Ronald Yang 2F

Also want to add that heat at constant pressure is not a state function just because it is equal to ΔH, which is a state function. Heat, under any condition, is a path function.

Ronald Yang 2F

Yes, a system not in equilibrium would undergo an irreversible process. Whether a process is reversible or irreversible refers to the state of the system at a certain point. A process is reversible when an infinitesimal change in the external pressure can reverse that process. For example, when a ga...

Ronald Yang 2F

When ΔT = 0, that doesn't mean q is also zero. In the example given today in lecture, for a reversible, isothermal expansion, heat must constantly be put into the system to keep the temperature constant, hence to keep ΔT = 0. The heat put in by the heat reservoir is used as work in expansion, so ΔV ...

Ronald Yang 2F

Hm...I'm not sure if how I reasoned it makes sense or is correct, but here: Imagine a gas inside a piston. Let's say initially it is at equilibrium, with external pressure being 2 atm and the pressure of the gas being 2 atm. For expansion to take place, there must be a difference in pressure between...

Ronald Yang 2F

You are right: if you do work on a system by compressing a piston, the energy would increase. w=-PΔV, so when a gas is compressed, ΔV would be negative, so overall, work would be positive. Since ΔU=q+w, assuming the system is under adiabatic conditions, internal energy would increase.

Ronald Yang 2F

Hey Kayti,

It's because the path heat takes does matter, since heat is a "transfer" of energy. It only exists during a change, not at the beginning and end of a change.

Ronald Yang 2F

The constant R is the same, whether you look at temperature in ^oCelsius or Kelvin. We are looking at a "change in temperature," and since the relationship is K = ^oC + 273.15, that change in temperature would be still be the same whether in Celsius or Kelvin.

Ronald Yang 2F

The easiest way is to look at the units for each constant. For 0.08206, that is in the unit (atm*L)/(mol*K). For 8.314, that is in the unit J/(mol*K). They are essentially the same thing other than the units. You can go from 0.08206 to 8.314 by using the conversion 101.325 J/(atm*L). Based on the in...

Ronald Yang 2F

Bond enthalpies refer to enthalpies for the breaking of certain bonds. When you calculate the enthalpy change for a reaction, you must add up the enthalpy changes due to the breaking of bonds in reactants and the forming of bonds in products. So, when we look at the reactants, we simply just add the...

Ronald Yang 2F

I would also like to add that, when steam comes into contact with skin, it immediately starts to condense. Since condensation entails a phase change from vapor to liquid, essentially hydrogen bonds are being formed, so an exothermic reaction takes place. The heat released in this phase change is tra...

Ronald Yang 2F

Enthalpy is a state function because it's value is determined solely by it's current state. It is not dependent on the path taken to reach that state, as enthalpy is dependent only on the initial and final conditions, which is why we usually compute ΔH. However, for something like heat, it's value i...

Ronald Yang 2F

Hey Jasmine! :) Increasing the oxidation state means a higher oxidation number on that Cl, which gives Cl more of a positive charge on that atom. Since like charges repel, the more-positive Cl would repel H+ more easily, making the overall molecule a stronger acid with higher H+ donating properties....

Ronald Yang 2F

I think that preference is stronger when NaOH is added to CH3COOH. For a buffer solution, both CH3COOH and CH3COO- are actually pretty weak, so rarely would either react with water. That's why they are good for making buffer solutions. They concern only themselves and not water (ex. strong acid adde...

Ronald Yang 2F

If no H values are given, the best way is to look at what's happening in the reaction. For the reaction N 2 O 4 <-> 2 NO 2 , a bond between the two nitrogens is broken apart. Since breaking a bond requires energy, this reaction must be endothermic. I'm now looking at the course reader, and this is p...

Ronald Yang 2F

Well, if it's an acid, most likely it won't be any kind of base. But, if you're confused about the difference between a Bronsted base and Lewis base or between a Bronsted acid and Lewis acid, Bronsted focuses on the transfer of protons (H+), while Lewis focuses on the transfer of electron pairs. A c...

Ronald Yang 2F

Technically, if K>1, there is a little bit more favorability toward the formation of products, since you can think of it as more products over reactants (the ratio is greater than 1). However, the general rule for determining the favorability of a reaction toward products or reactants is as Hank say...

Ronald Yang 2F

If an acid is a strong acid, it can easily donate a proton (Ka is big, since products favored). Thus, when you look at it's conjugate base, the conjugate base would not likely accept a proton, as it wouldn't want the H+ back that it so readily gave up. Think of it this way. An acid would technically...

Ronald Yang 2F

I had a question about sig figs for equilibrium problems. On paper, when I work with the quadratic part of the problems, when I multiply out terms, do I write numbers based on the sig figs of the numbers I multiplied but use the longer numbers to calculate change in x or do I write them all out on p...

Ronald Yang 2F

Oh, I understand now. I feel like the reason the manual did that was that it could use an approximation of zero to estimate x, since K is so small. If you just did the regular way, each equilibrium concentration would have a number "+x" or "-x" within it. When you push the reacti...

Ronald Yang 2F

There are three initial pressures. The 1.0 mol of H2 refers to H2 when it is at equilibrium. If you are confused about how to proceed, you would find the pressure of H2 at equilibrium and use PV=nRT to figure out the volume of H2 at equilibrium.

Ronald Yang 2F

If the volume of the container decreases, essentially pressure in the container increases (pressure and volume have an inverse relationship). This pressure is created by the collisions of gas molecules. Thus, we only look at molecules in the gas phase. For 2HCl(g) + I(s) --> 2HI(g) + Cl2(g), we only...

Ronald Yang 2F

I'm having trouble understanding what you're trying to ask, but pi bonds don't utilize hybridized orbitals. They use unhybridized orbitals. For example, if you look at a C atom in ethene, the Carbon will use sp 2 hybridized orbitals to form sigma bonds, but for that pi bond with the other Carbon, it...

Ronald Yang 2F

You are correct. Here's an older discussion about this that I just found by looking this up. viewtopic.php?f=45&t=2356

Ronald Yang 2F

The "bis" in "chlorobis" is probably describing some other ligand rather than Cl. Also, you would use the "bis" and "tris" kind of prefixes for polydentate ligands as well.

Ronald Yang 2F

I guess for HN(CH2CH2NH2)2 looking at the lone pairs could work, since only at the Nitrogens can it create coordinate covalent bonds with a transition metal. But, the thing is, a molecule may have an atom with lone pairs, but that atom may not form a coordinate covalent bond with a transition metal....

Ronald Yang 2F

Actually, the fact that it is potassium hexacyanochromate(III) means that the potassium has a positive charge (+1). The fact that Chromium has a 3+ oxidation state does not make the whole hexacyanochromate(III) complex have a net positive charge. Also, the fact that the complex includes "chroma...

Ronald Yang 2F

Well, you can infer from a molecule's shape if it's a lewis base. Usually, an atom, ion, or molecule with a lone pair of electrons can be a Lewis base, since it can donate that electron pair. Thus, if it has a shape like trigonal pyramidal, bent, or other shapes that include lone pairs and follows a...

Ronald Yang 2F

They are all isoelectronic, so basically the one with a lower effective nuclear charge would have a higher ionic radius (essentially, the one with less protons). Thus, in increasing ionic radius form, that would be Cl^-<S^2-<P^3-.

Ronald Yang 2F

Well, usually the most stable structure would be the one with zero formal charge on every atom, which means the central atom would have the same number of electrons closest to it as the number of valence electrons. For example, SF6 can make 6 bonds because it has 6 valence electrons. XeF2 makes 2 bo...

Ronald Yang 2F

If you think about it, even if in the lewis structure the Cl's are opposite with regard to the Carbon or on the same side, they still are oriented as a Cl-C-Cl bond in reality. No matter how you place the Cl in a lewis structure, if you were to translate that into a tetrahedral 3D arrangement, the m...

Ronald Yang 2F

I thought that lone pairs affect the shape of every molecule, including the linear ones (ex. bent) You are correct; lone pairs do affect the shape of a molecule. So, for a molecule to be 180 degrees, it can have no lone pairs, such as BeCl 2 . But, a molecule can also have lone pairs and still be 1...

Ronald Yang 2F

There is no shape for describing an entire molecule with two central atoms. However, you can describe the shape at a single atom. For example, for cis-dichloroethene, at either Carbon atom, we can determine that it's shape is trigonal planar. There won't be an overall shape like bi-trigonal planar o...

Ronald Yang 2F

Jennifer Cheng 4H wrote:The trigonal pyramidal structure actually has four areas of electron density.

You are totally correct, but don't mix up trigonal pyramidal with trigonal bipyramidal.

Ronald Yang 2F

For elements that can have an expanded octet, usually the max number of valence electrons the central atom can take on is the amount of its outer valence electrons. For example, in PCl 5 , phosphorus has 5 valence electrons, so usually, it can bond with at most 5 other atoms. Again, in XeF 6 , xenon...

Ronald Yang 2F

Trigonal bipyramidal and seesaw shapes have the same number of regions of electron density, as you said: 5. If you were to look at PCl 5 and SF 4 , both have an electron shape of trigonal bipyramidal, as they both have 5 regions of electron density. However, PCl 5 has no lone pairs, whereas SF 4 has...

Ronald Yang 2F

I think you meant 3 "double" bonds, but, the reason why oxygen with the -1 is the solution is because oxygen is more electronegative than chlorine. Chlorine is not as electronegative, even though it is to the right of oxygen (it still is a row below). Putting a formal charge of -1 on chlor...

Ronald Yang 2F

First, start off with Heisenberg's Indeterminacy Equation: \Delta p \Delta x\geq \frac{h}{4\pi } . You can break down Δp into mΔv, since mass is constant, so m\Delta v \Delta x\geq \frac{h}{4\pi } . Δx is what you want to find. You can find Δv, which is uncertainty in velocity, by taking the +/- 5 m...

Ronald Yang 2F

Chromium and Copper can obtain a more stable electron configuration when they take an electron from the 4s and add it to 3d. When it does this, it becomes either a half-filled shell (Cr) or a full one (Cu). This results in a more stable compound with lower energy. Half-filled shells and full shells ...

Ronald Yang 2F

I don't think it is required to memorize that Joules is also kg*m^2*s^-2. However, there are times when it would help to know these alternative units to figure out the answer. For example, for Debroglie's wavelength, to find the wavelength of an "object," you would need to know the mass an...

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