CHEMISTRY COMMUNITY

ryanhon2H

When writing the rate law using pre-equilibrium, does it matter if we write the answer as k₂K or k₂k₁/k'₁?

ryanhon2H

If you subtract cathode - anode, you don't flip the signs. If you add them, then you would have to flip the sign of the anode.

ryanhon2H

Yes, it is relevant when balancing the charges of the equation.

ryanhon2H

In 6th edition 15.17 it involves a zero order reactant.

ryanhon2H

A zero order reaction is one where the rate is independent of the concentration. An example could be the decomposition of nitrous oxide on a hot platinum wire. The rate of that reaction doesn't depend on the concentration of nitrous oxide, it depends on whether surface area runs out on the wire for ...

ryanhon2H

I think it is because H20 is a liquid, so you don't include the concentration of it.

ryanhon2H

You have to use the equation ∆G = -nFE. Looking at that relationship, you can see that when E is positive, ∆G is negative, and when E is negative, ∆G is positive. A reaction is spontaneous when ∆G is negative, and thus when E is positive.

ryanhon2H

I know that in writing a cell diagram, the anode side goes first and then the cathode side goes next. But is there a way to know the order to write the terms of the anode and cathode? Like solids first then aqueous? Or does it not matter?

ryanhon2H

Platinum is used when there isn't a conducting solid. So in the textbook, there can be platinum on a side with a solid because that solid isn't a conductor. Platinum is used because it is inert and won't react.

ryanhon2H

An inert conductor is used if there is no metal electrode. In this case copper is metal and an electrode, so an inert conductor would not be needed.

ryanhon2H

Inert conductors transfer electrons, so in redox reactions they are the medium for the oxidation and reduction half-reactions.

ryanhon2H

Since electrons are moving from the anode to the cathode, the salt bridge balances out the charges by donating negatively charged anions to the positively charged anode, and positively charged cations to the negatively charged cathode.

ryanhon2H

Because entropy is a state function, the change will be the same regardless of whether the reaction is reversible or not, since it is not path dependent, so you can use q rev /T, as long as the reaction is isothermal. If it is isobaric (constant pressure), isochoric (constant volume), etc you would ...

ryanhon2H

Negative ∆G means the forward reaction is spontaneous. Positive ∆G means the forward reaction is not spontaneous, but the reverse reaction is. If ∆G equals 0, that means the system is in equilibrium.

ryanhon2H

Gibbs free energy is given by the equation ∆G = ∆H - T∆S An exothermic reaction is much more likely to be spontaneous than an endothermic reaction, as it releases heat instead of requiring it. The ∆H of an exothermic and likely spontaneous reaction is negative. A reaction that also increases entropy...

ryanhon2H

For bond enthalpies, the C-C and C=C bonds of benzene are counted the same. So since benzene has 3 single carbon bonds and 3 double carbon bonds, it is a total of 6 C-C/C=C bonds formed.

ryanhon2H

The copper is losing heat to the water, so the equations would be set equal to each other. However, you would have to flip the sign of the copper because the heat lost would be negative, so you get -q copper = q water By using (100-T) you get rid of the negative sign so both values become positive. ...

ryanhon2H

You can determine degeneracy by looking at all the possible arrangements of the molecules, like if the positions of atoms changed.

ryanhon2H

The first equation is used when the reaction is an irreversible expansion against constant pressure. The second equation is used when the reaction is a reversible expansion.

ryanhon2H

Work can be negative if it is done by the system, opposite of doing work on the system in that heat is released in contrast to adding heat, or energy.

ryanhon2H

Isothermal is when temperature is constant. I think isometric is related to isomers, where molecules that can be superimposed after a reflection like in a mirror are called isomers.

ryanhon2H

Endothermic is positive ∆H because the enthalpy of the products is greater than the enthalpy of the reactants, so energy is absorbed. The opposite is true for exothermic, since energy is released.

ryanhon2H

Heat is dependent on work due to the first law of thermodynamics, where ∆U = q + w, so q = ∆U - w. Work is path dependent since the steps taken to get from initial to final conditions can differ.

ryanhon2H

Heat is a transfer of energy. Enthalpy is a state function and cannot be directly measured.

ryanhon2H

The first explanation is correct. You are right in saying that water has no effect on the equilibrium constant. However, the question is asking about the equilibrium compositions, or concentrations. For this, I don't think you can assume water is in excess. Thus, when more water is added, the aqueou...

ryanhon2H

I think K is referring to Kp, so it would just be Kp and Kc.

ryanhon2H

This was in Chem14A, where some anions or cations are basic/acidic when dissociating in water so they can affect the pH. On page 478 in the 7th edition textbook there are a couple of tables that say which ions affect pH.

ryanhon2H

No, since equilibrium constants are measuring concentrations or partial pressures, and the concentrations or pressure of solids and liquids are practically constant so they don't affect the concentration for ions or gases.

ryanhon2H

By taking away the product, the reaction is no longer in equilibrium, so more product will be produced to replace the amount taken away. It's more product yield in the sense that more product is being made, not necessarily more total product.

ryanhon2H

I think Dr. Lavelle said in class that pressure can also be changed by pumping an inert gas, so that wouldn't change the concentration compared to changing the volume, but K would still stay the same either way.

ryanhon2H

You can tell if a molecule is polydentate if there are multiple atoms in a molecule that have lone pairs.

ryanhon2H

Yes, but generally the question will ask you to find the minimum value so you can just use the equal sign in that case.

ryanhon2H

I understand that hydrogen bonding occurs between hydrogen and N, O, F, but I'm confused why the hydrogen atoms on a molecule aren't counted as being able to form hydrogen bonds. Can't those hydrogen form hydrogen bonds with N, O, or F on another molecule? For example, a terminal oxygen atom is coun...

ryanhon2H

Strong bases fully dissociate in water, while weak bases only partially dissociate, as the strength of the base depends on its ability to fully dissociate.

ryanhon2H

For the acids, there are in total 7 strong acids that you should just memorize: HCl, HBr, HI, HNO3, HClO3, HClO4, H2SO4. (so the rest would be weak acids). The strong bases, instead you can recognize them as they are always a metal hydroxide (e.g. LiOH, NaOH, KOH...). For the metal hydroxides (stro...

ryanhon2H

Amines are basically any compound that has nitrogen with a lone pair, where it's derived from ammonia by replacing one or more of the hydrogen atoms.

For example, using R to stand in for the replacement groups, amines could be

NH₃ ; NH₂R ; NHR₂

ryanhon2H

Xenon is in Period 5 which is after Period 3, so it can be an exception to the octet rule and have more regions of electron density, which allows it to have 6 regions of electron density for XeF₄

ryanhon2H

You can use the VSEPR model to help understand this. CH 2 2- has 2 bonding pairs (the 2 H atoms) and 2 lone pairs. The VSEPR formula then is AX 2 E 2 . Based on the formula, the molecular shape is bent. This is similar to the molecular geometry for water. The lone pairs exert greater repulsion compa...

ryanhon2H

ClF3 has 2 lone pairs and its molecular shape is t-shaped. Because of the 2 lone pairs, the dipoles don't cancel, so ClF3 is polar.

ryanhon2H

I think what it means is that VSEPR models can tell you the general shape of the molecule, such as bent, tetrahedral, square planar, etc, so that would be qualitative. Quantitative I think is the the angles of the bonds, so while VSEPR models can predict the general molecular geometry, it can't pred...

ryanhon2H

Electron geometry is the geometry of all the electron groups, including lone pairs and bonds. Molecular shape is the actual position of the atoms, not including lone pairs. For example, NH 3 electron geometry is tetrahedral (due to 3 bonds and 1 lone pair), but the molecular geometry is trigonal pyr...

ryanhon2H

I think that molecular shape is important in relation to biology, for example proteins get their function from their shape, so that's why the actual physical shape is more important than the electron geometry.

ryanhon2H

Polarizability can be seen as trending the opposite way of electronegativity. Polarizability is how easy the electrons of an atom/molecule are distorted. If an atom has a high electronegativity, they have a stronger hold on their electrons, and thus the electrons are not easily distorted. If an atom...

ryanhon2H

The way to figure this out is using VSEPR. Basically you can use the Lewis structure to figure out the molecular geometry, using the VSEPR notation of AX n E y , where A represents the central atom, X represents the atoms bonded to the central atom, n is the number of bonds to the central atom (doub...

ryanhon2H

Just remember that the shorter the bond, the stronger it is. Cl is bigger than F, so the bonds of C-Cl will be further apart than C-F, and so C-Cl bonds will be weaker than C-F. Br is bigger than Cl, so C-Br bonds will be further than C-Cl bonds, so C-Br will be weaker than C-Cl. C-F bonds are the s...

ryanhon2H

I believe that for elements with more than one electron, so basically every element except hydrogen, degeneracy is determined when the values for n and the values for l are the same.

ryanhon2H

In class professor Lavelle did an example where he drew the SO4^2- ion with all single bonds, and then drew another structure where there were 2 double bonds. Which one would be the correct way to draw it? Would both technically be correct, just that the second one is more correct due to the formal ...

ryanhon2H

I think you just forgot to solve for ∆p. You solved only for momentum (p) but just forgot the step to get ∆p. The question tells you that ∆p is (1x10^-6) times the momentum, you have to solve for p first using p = mv, then multiply the result by (1x10^-6). I think you just forgot that last part. Wha...

ryanhon2H

I know that electron affinity is the energy released to add an electron to an atom. I'm just a little confused on why energy is released? Why wouldn't energy be required to add an electron?

ryanhon2H

One way to figure the questions out is to just add up the number of electrons. For example, 2A.9 a) is [Ar]3d7. Ar has 18 electrons, plus the 7 given in 3d7, for a total of 25 electrons. The question tells you that it is a +2 charge ion, which means the neutral atom lost 2 electrons. 25 electrons is...

ryanhon2H

Valence electrons are electrons that are in the outermost shell for an atom. The outermost shell for Sb then would be the 5p orbital, so the valence electrons would be the electrons in 5p, which is 5. It wouldn't include the d-orbital since the d-orbital is not the outermost shell.

ryanhon2H

h bar is h/2π. The difference is in the Heisenberg equation where it could be (Δp x Δx) = h/4π or (Δp x Δx) = 1/2(h bar). 1/2(h bar) simplifies to h/4π

ryanhon2H

Henry_Phan_3B wrote:
Chem_Mod wrote:Yes, E = hv or E = hc/ $\lambda$ .

How did you get that second equation?

You substitute for v in E=hv to get E=hc/λ

Use the light equation c = vλ, solve for v to get v = c/λ, and substitute back into E=hv for v

ryanhon2H

Is there a specific order for values to be assigned to Px, Py, Pz? In lecture I remember he said it was arbitrary and that it didn't matter? I'm a little confused on that.

ryanhon2H

h bar is said in the textbook to be h/2π, where h is Planck's constant The uncertainty principle equation is (∆p)(∆x) = .5(h bar) or (∆p)(∆x) = h/4π ∆p is the momentum uncertainty and ∆x is the position uncertainty. ∆x is given in the problem, which is 350. pm or 3.50 x 10 -10 meters. Plug it into t...

ryanhon2H

The different series correspond to different parts of the hydrogen spectrum. For example, the Balmer series is used for the visible light spectrum of hydrogen, while the Lyman series is used for the ultraviolet spectrum of hydrogen. As Atul said, the series have different base energy levels. This is...

ryanhon2H

The way I did it was using the Rydberg equation, which is 1/wavelength (in meters) = R(1/n 1 2 - 1/n 2 2 ) The wavelength is given, and R is the Rydberg constant. Since this is in the ultraviolet spectrum of hydrogen, it is the Lyman series, where n 1 is 1. Now all you need to do is plug in and solv...

ryanhon2H

I think the safest option is to just round to however many significant figures were given in the problem.

ryanhon2H

For part C on M11 (7th edition), the answer in the back is given only 2 sig figs as 5.7g. The answer I got was rounded to 3 sig figs, at 5.79g. I was just wondering it's supposed to be 3 sig figs, as the answer to part B is 3 sig figs, or if my answer is wrong? Since if I round it to 2 sig figs it w...

ryanhon2H

O2 is the limiting reactant for part a. You first have to figure out the limiting reactant for the first reaction using mole ratios, which turns out to be P4. This means that all of the P4 is used in the first reaction, but only some of the O2 is used. Since you now know that P4 is the limiting reac...

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