CHEMISTRY COMMUNITY

Tiffany_Chen 2K

Intermediates will not show up in the final product, catalysts will as it is not degraded.

Tiffany_Chen 2K

Thank you for an awesome two quarters! Best of luck in your future endeavors :)

Tiffany_Chen 2K

Lavelle should update us soon (hopefully). Might be an online test, or we have to scan it in by a certain time?

Tiffany_Chen 2K

Your TA should email you with details!

Tiffany_Chen 2K

Yesss, we love a student-run review session! ^Trustworthy source :)

Tiffany_Chen 2K

Yes and electrons will flow from anode to cathode! Thus, a positive cell potential for reaction to be spontaneous.

Tiffany_Chen 2K

NO3 is generally used as a salt bridge. OH- is added when the redox is in a basic solution (to balance oxygens).

Tiffany_Chen 2K

Additionally, cell potential would be the maximum energy available to do work.

Tiffany_Chen 2K

Oxygen has a -2 charge for each atom, given a total of -8 charge. Since the overall molecule is -1, Mn will have to be +7.

Tiffany_Chen 2K

Ruby Richter 2L wrote:When adding reactions and in the final state how do you determine what physical state everything is in?

You should be able to tell the states given the half reactions. Soluble compounds can also form (from reactant solids).

Tiffany_Chen 2K

Think of it like this: for the reducing agent (so the reactant is being oxidized), it will be positively charged after losing electrons. Thus, you can add OH- to balance out the positive charge.

Tiffany_Chen 2K

Either would work, log is typically used as it is more useful in calculating pH.

Tiffany_Chen 2K

The more positive E(red) will be at the cathode, so you can find E(ox) of the other half reaction by reversing the given E(red) value. Then add them to find the overall reaction cell potential.

Tiffany_Chen 2K

Last page of thermochemistry, and all of electrochemistry!

Tiffany_Chen 2K

Oxidation number would be 0 as the overall charge of O3 is 0.

Tiffany_Chen 2K

If the question asks for the answer to be in specific units, don't forget to convert at the end. For thermodynamics, usually celsius/kelvin can be interchangeable due to the same difference.

Tiffany_Chen 2K

Anode: site of oxidation, cathode: site of reduction.

Tiffany_Chen 2K

How do you find oxidation numbers? Group one would be +1, group two would be +2, group seventeen would be -1. Neutral compounds would have a total oxidation number of 0, so you can usually figure out individual oxidation numbers of atoms from knowing one atom out of the compound (E.g. CrCl3, Cl has...

Tiffany_Chen 2K

Yep, as no heat transfer will occur (isolated system).

Tiffany_Chen 2K

Entropy in the natural world is always increasing, if the reaction is reversible, entropy would then equal 0.

Tiffany_Chen 2K

At equilibrium, deltaS(total)=0. Thus, deltaS(sys) will have the opposite relationship to deltaS(surr). Any entropy change will be the only entropy change to the system (as deltaS(surr) will only reflect the exact opposite of deltaS(sys)).

Tiffany_Chen 2K

In the opposite way of thinking, reversible expansions are where there is an infinitesimal amount of changes to the system, thus allowing the system to be reversible. Irreversible expansions are typically a huge, sudden change that will take an equally large amount of energy to reverse (which is mos...

Tiffany_Chen 2K

Adding on, for a regular calorimeter, it is constant pressure (air can diffuse in and out). Bomb calorimeter, it is constant volume as it's a sealed container, thus you would use Cv.

Tiffany_Chen 2K

Just make sure the constants you use (e.g. R), will match the units you want. Really depends on what the question asks for.

Tiffany_Chen 2K

When deltaG (free energy available to do work) is negative!

Tiffany_Chen 2K

For the third law, "entropy reaches a constant value (0 for perfect crystals) when temperature approaches 0 K". 0 K would indicate the material is only present in one possible micro-state/position.

Tiffany_Chen 2K

Think of area under the curve, reversible work will have a gradual decrease in temperature, giving more area. For irreversible reactions (where pressure is constant after a force is exerted), the area will be just a box (recall the diagram from class).

Tiffany_Chen 2K

Heat for a reversible equation is qrev. Reversible equations are when the equation is at equilibrium.

Tiffany_Chen 2K

Yep, in terms of molecular states the Boltzmann equation will be useful for calculating entropy (statistical). The other equation for entropy is good for thermodynamics as a whole.

Tiffany_Chen 2K

Really depends on specific values (using the equation). We should cover the +/- conditions post midterms. For entropy, you can think of the second law, where spontaneity and increase in entropy relates (-deltaG corresponds with spontaneous reactions).

Tiffany_Chen 2K

For very strong acids, their pH will be extremely low and can thus create a negative outcome using the log equation.

Tiffany_Chen 2K

For problems, inert gases will not show up in the equation (thus adding it will not shift equilibrium).

Tiffany_Chen 2K

Any gas that exhibits intermolecular forces and takes up volume.

Tiffany_Chen 2K

A reversible reaction would not increase the entropy of the system or surroundings, as it maintains a thermodynamic equilibrium. A irreversible system is most common, as it takes the reaction away from equilibrium (then energy will need to be put in to reverse the system).

Tiffany_Chen 2K

Agreeing with the above comments, here is also the proof:

K=C+273.15
K1-K2=C+273.15-(C+273.15)
K1-K2=C1-C2

Tiffany_Chen 2K

In addition, remember that only temperature will change the reaction's equilibrium constant. A catalyst will only speed up the reaction for equilibrium to be reached more quickly.

Tiffany_Chen 2K

No shift in reaction will occur if there are equal moles on both sides.

Tiffany_Chen 2K

Chemical equilibriums will shift accordingly to minimize stressors. In lecture, we covered stressors such as concentration, pressure, and heat.

Tiffany_Chen 2K

Both, if the ionization percentage is less than 5%, this justifies in approximating x as 0.

Tiffany_Chen 2K

For your final answer (if asked for pH), you want to have the least number of sig figs for the value past the decimal point. E.g. you are given 0.015M of something and that is the least number of sig figs. Your answer would be need two sig figs past the decimal point, e.g. 3.00

Tiffany_Chen 2K

I would just to be safe! You can check using percent ionization too, if the percent ionization is <5%, then x approximation is justified.

Tiffany_Chen 2K

Look at PV=nRT but rearranged into P=((n/V)*RT) or P=conc*RT. For example, volume is halfed ->> n will double. You can then recalculated Qc and compare this to Kc to determine which way the reaction will go.

Tiffany_Chen 2K

The question should indicate which one to use. Though all reactants/products can be in gas form, we can still look and compare their concentrations (Kc).

Tiffany_Chen 2K

Reaction quotient is used to determine whether the equation is at equilibrium (so compare to the K value). If not, we can predict which direction the reaction needs to favor in order to reach the equilibrium rate for both sides.

Tiffany_Chen 2K

Yes, increasing temperature will increase the value of Kw as the autoprotolysis of water is endothermic (so increasing heat will lead to formation of more products).

Tiffany_Chen 2K

If you think of the reverse reaction (2H2+O2->2H2O), energy is released to form the covalent bonds of water. Thus the reverse is exothermic, while the forward reaction (protonation of H2O) is endothermic.

Tiffany_Chen 2K

Solids and liquids will hold relatively constant volume, thus you can technically "cancel" them out when putting products over reactants to find Kc.

Tiffany_Chen 2K

Kc is used for concentrations (M), which can be gases or aqueous molecules, Kp uses partial pressures of gases. Question should be clear on which one to use.

Tiffany_Chen 2K

If K=10^-3 or greater, full calculations using the quadratic equation should be used

Tiffany_Chen 2K

Anything gas or aqueous should be included. Solvents (l) or solids will not be including in calculating Kc/Kp.

Tiffany_Chen 2K

Yes, using variables will aid you in calculating different parts of the Kc equilibrium. You will use the quadratic equation to solve for the 'x', then apply to the reactant or product.

Tiffany_Chen 2K

For #31 should the name be ...nickel(III) ion? The complex has a 2+ charge.

Tiffany_Chen 2K

Nyari Muchaka_Discussion 4A wrote:pH cannot be negative, it exists between 0-14.

But taking the -log[1.5]=-0.176, do we just assume it as positive? Or basically very close to 0 I guess.

Tiffany_Chen 2K

For d), oxalate is bidentate as there will be two double bonds (these won't bind to the metal as the pi bonds restrict rotation, thus only 2 of the single bonded oxygens can bind at once).

Tiffany_Chen 2K

Resonance structure, one structure has two double bonds (leading to monodentate), one structure has one double bond (so two available oxygens, bidentate).

Tiffany_Chen 2K

But the book converts the -0.176 ph into a positive number. Why is that?

Tiffany_Chen 2K

CH3COO- is the conjugate base of acetic acid (as it is a proton acceptor (takes in H+ from H2O, leaving OH-, which creates the basic solution)).

Tiffany_Chen 2K

Metal oxides generally form basic oxides, while nonmetal oxides generally form acidic oxides. Sulfur is a nonmetal thus will be acidic. Amphoteric oxides (shown in class) can react as both an acid or base to form a salt and water.

Tiffany_Chen 2K

I've seen EDTA under two names - ethylenediaminetetra acetic acid and ethylenediaminetetra acetato . Which one is correct? Additionally, I've seen structures with hydrogens attached to the acetate ends and structures without (so there is a negative charge)––all under the name of EDTA. I thought acet...

Tiffany_Chen 2K

You would also use the prefix if the ligand is polydentate, and there are more than one of the ligands (so no di-dien).

Tiffany_Chen 2K

Yes, the first bond would be a sigma bond, followed by a pi bond for each additional bond. For example, for a sp hybridization (triple bond), there would be one sigma bond and two pi bonds (using the two empty 2p orbitals).

Tiffany_Chen 2K

Polydentate means the number of donor groups a ligand can bound to the transition metal. E.g. EDTA is hexadentate, as there are 4 oxygens and 2 nitrogens available to bind to the TM.

Tiffany_Chen 2K

Yes! Also recall ionization energy is the energy need to remove a valence electron, and electron affinity is the energy released when an electron is bound to a neutral atom in gas phase.

Tiffany_Chen 2K

I found this post to be helpful (viewtopic.php?t=764). I believe if the ligand is polydentate and will form a "claw" structure when binding to the same transition metal, it can be categorized as a chelate.

Tiffany_Chen 2K

Think of the bonding regions as regions of electron density (triple/double bonds are each considered one region, same for single bonds).

Tiffany_Chen 2K

No calculations, but understand the concept, which is that greater distance will lead to a PE of 0 (no interaction), and greater polarizability of the two atoms will lead to a more negative PE (as there is a negative sign in front of the equation, and negative PE is favored).

Tiffany_Chen 2K

Yes! The bond angles between the equatorial atoms will actually be less than 120 degrees, as well as less than 90 degree angle between the axial atoms and equatorial atoms.

Tiffany_Chen 2K

For example, in carbon, hybridization would occur in 2s/2p orbitals (as these electrons are used in creating hybrid orbitals). Both 2sp or sp would work (2sp is just more specific as it gives the quantum number).

Tiffany_Chen 2K

I think because oxygen is much more electronegative than Se, it exerts a greater pull on the electrons and pulls out. The dipoles may not cancel out because there are multiple resonances of H2SeO4. For example, if the OH and O are not symmetrically bonded to the Se (like an O across and O and an OH...

Tiffany_Chen 2K

Why would there be dipole forces for molecules of H2 and SeO4? Aren't they both nonpolar?

Update: the question asks for the entire molecule of H2SeO4, makes more sense!

Tiffany_Chen 2K

The shape would be tetrahedral - try drawing it out with different atoms on top (the polar vectors would not cancel due to the shape, no matter if H or Cl is the top atom).

Tiffany_Chen 2K

Definitely get started on homework from Outline 4, also review the intermolecular forces.

Tiffany_Chen 2K

Can you give an example? It depends on the formal charge/molecule.

Tiffany_Chen 2K

The molecular shape for N2H4 would be trigonal pyramidal as it is A34E1. The general rule for that shape is that bond angles are less than 109.5, 107 is exact and obtained from experimental data (so less than 109.5 degrees would work as the answer, the book is just exact).

Tiffany_Chen 2K

I would say that's the most efficient and clearest way to do limiting reactant problems for this class.

Tiffany_Chen 2K

Chlorine is more electronegative than iodine (atomic trends). The difference in electronegativity between HCl will then be greater than HI, therefore ionic character is more expressed in HCl.

Tiffany_Chen 2K

I guess that would qualify as resonance (https://commons.wikimedia.org/wiki/File ... nce-2D.png), but in this case we are looking for radicals, and the structure itself (either one) would be stable.

Tiffany_Chen 2K

Yes, it refers to threshold energy (energy to remove the e- from the metal surface).

Tiffany_Chen 2K

I think it has to do with HClO3 being a strong acid, thus ClO3 would be a structure, and the remaining hydrogen would be bonded to an oxygen (as H+ will dissociate in water to form H3O+, leaving ClO3-).

Tiffany_Chen 2K

Do we have lecture on Wednesday?

Tiffany_Chen 2K

There should be a couple more review sessions, including an in-depth one on Sunday.

Tiffany_Chen 2K

You would take the number of bonds of that molecule (after drawing the lewis structure) divided by the number of bond groups (e.g. NO3 would have 4 bonds/3 bond groups = bond order of 1.33). The higher the bond order means the electrons are more tightly attracted, thus a greater stability.

Tiffany_Chen 2K

Resonance structure is where the placement of the double or triple bond alternates, while a resonance hybrid structure would take in the average of the resonance structures for that molecule to display the a more realistic electronic structure.

Tiffany_Chen 2K

So Cu is one of the d-block exceptions, where it would be more stable (symmetric distribution of electrons) to fill all of the 3d block (as 4s is technically higher in energy) first, before moving on to the next energy level.

Tiffany_Chen 2K

I would say perform formal charge calculations to structures where you are uncertain of the middle atom (such as OHCl, where formal charges can determine whether O or Cl would be in the middle).

Tiffany_Chen 2K

Does he have modules for each unit/focus?

Tiffany_Chen 2K

There are definitely exceptions further along the periodic table, but as long as you follow the general trend/reasoning used for upper elements you should be good (exceptions except for those mentioned in class should not be tested).

Tiffany_Chen 2K

For atomic radius: let's say we are going from right to left across a period. The effective nuclear charge would increase, as the number of protons increase but the number of inner electrons stay the same. Thus, the nucleus will have a higher hold on the outer electrons, decreasing the atom's radius.

Tiffany_Chen 2K

Ionization energy is the energy required to remove an electron from an atom in gas phase. Going across a period, you are increasing the number of photos while the number of shielding electrons (inner electrons) stay constant. So there will be an increase of effective nuclear charge, a tighter hold o...

Tiffany_Chen 2K

I believe the electron will absorb the exact needed energy to go up an energy level, and the excess energy will be released.

Tiffany_Chen 2K

Arianna Perea 3H wrote:What happens if the uncertainty is less than h/4pi?

The best scenario would be equal to h/4pi.

Tiffany_Chen 2K

Avogadro’s number can be towards any unit, as long as it’s represented as 6.022*10^23unit/mol.

Tiffany_Chen 2K

Yes, and we can use the speed of light as a basis to compare our calculations to.

Tiffany_Chen 2K

Hey just asking what the wave functions represent exactly, and will we have to know actual numeric values for the functions? They represent a mathematical probability of finding electrons. Just know that the (wave function)^2 gives you a positive graph, with peaks being places with high electron de...

Tiffany_Chen 2K

905385366 wrote:I was looking at the solutions manual and the value for h is not planks constant (6.626x10-34) but some other value. Can somebody explain why it is different?

The value given should be h/4pi, which is used in Heisenberg's indeterminacy equation.

Tiffany_Chen 2K

Intensity of the light relates back to how scientists previously thought light could only behave as a wave, thus the thinking of how increasing the intensity of the light (wave amplitude) would increase the energy needed to eject the electrons. However, this model failed and led to the particle mode...

Tiffany_Chen 2K

Try to take at least 2 or 3 values past the decimal point when using molar mass of elements. Also try not to round until the final answer to be exact as possible.

Tiffany_Chen 2K

You can use the uncertainty in velocity that was found and plug into the E=1/2mv^2 to find the uncertainty in kinetic energy. The units for that would be J/e, so you can then convert to moles by multiplying Avogadro's number.

Tiffany_Chen 2K

Hydrogen only has one electron, as compared to other elements where removal of one electron will lead to electron-electron repulsions (likewise the Rydberg equation only works for Hydrogen). I think Lavelle was describing how you can use other elements if you remove all but one electron.

Tiffany_Chen 2K

Just make sure to bring a non-graphing calculator!

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