CHEMISTRY COMMUNITY

Sam Wentzel 1F 14B

How do we solve B)?

Sam Wentzel 1F 14B

E* = 3.03 V for F2(g) + 2 H+(AQ) + 2 e- -> 2 HF (aq), E* = 3.03 V Find Ka of HF the E* value for F2 + 2 e- -> 2 F- is 2.87 V. If we want the cell to be favorable, we want a positive E*cell value right? So why in the solutions manual do they do E*cell = 2.87 V - 3.03 V? = -0.16 V (Why do they choose ...

Sam Wentzel 1F 14B

(a) If you were to construct a concentration cell in which one half-cell contains 1.0 M CrCl 3aq) and the other half-cell contains 0.0010 M CrCl 3aq), and both electrodes were chromium, at which electrode would reduction be spontaneous? How will each of the following changes affect the cell potentia...

Sam Wentzel 1F 14B

a) Write balanced half-reactions for the redox rxn of an acidified solution of potassium permanganate and iron(II) chloride. b) Write the balanced equation for the cell reaction and devise a galvanic cell to study the reaction by writing its diagram. How do we identify which ions are spectator ions ...

Sam Wentzel 1F 14B

I also am very confused about the difference between the half reactions and the cell reaction?

Sam Wentzel 1F 14B

6L.7: Write the half-reactions and devise a galvanic cell (Write a cell diagram) to study each of the following reactions: a) AgBr(s) -> <- Ag+ (aq) + Br- (aq) b) H+(aq)+OH-(aq)->H2O(l_ c) Cd(s) + 2 Ni(OH)3(s) --> CD (OH)2(s) + 2 Ni(OH)2 I can break up the reactions into their half reactions and bal...

Sam Wentzel 1F 14B

1st order reactants will increase / decrease rate of reaction directly proportional to the increase / decrease in concentration of that reactant. 2nd order reactants will decrease the rate of reaction proportional to any increase in concentration of that second order reactant. 0 order reactants wont...

Sam Wentzel 1F 14B

We know that rate law is determined by the slow portion of the mechanism. Initial proposed rate law = k[I-][HCLO] But we know this is not correct because [HCLO] is an intermediate (produced, then consumed in rxn) and cannot participate in the rate law of the reaction. Therefore we need to solve for ...

Sam Wentzel 1F 14B

You need to use the Arrhenius Equation.

k' = k of reverse rxn
k = k of forward rxn
K = Equilibrium constant

k'/k = K

Hope this helps!

Sam Wentzel 1F 14B

The salt bridge adds opposite ions to the anode and cathode basically to keep their solutions from getting "too positive" and "too negative" respectively. For the anode, the salt bridge gives anions to replace lost e-'s. For the cathode, the salt bridge gives cations to replace l...

Sam Wentzel 1F 14B

They can't be used interchangeably. Use log for Ecell = E*-0.0592/n*logQ (Aka, when system is at 25 * C)

Use Ecell = E*-RT/(nF)lnQ for all other temps. Hope this helps!

Sam Wentzel 1F 14B

I know that for reactions in basic solution, I must balance the final H+'s of the reaction with OH-'s.

I was wondering what ways I can tell the reaction occurs in basic or acidic solution?

Sam Wentzel 1F 14B

E = E*-RT/(nF) x lnQ Calculate E* Using E* = E*cat-E*anode (substance w/ higher E* value from the E* values table is the cathode, substance w/ lower is anode) Determine number of moles of e-'s transferred by looking at balanced reaction, and analyzing Oxidation numbers. Here we find that n = 2. Fina...

Sam Wentzel 1F 14B

Q is heat transfer from a system, H is energy transfer from one side of a reaction to another

Sam Wentzel 1F 14B

(X)|(Compound)|(A)||(A)|(Compound)|(Y)
^ ^ (XA) ^anion ^(YA) ^cathode
| Anode

Sam Wentzel 1F 14B

You have to memorize them for O and H, but I believe the other elements can usually be inferred. O: -2 unless attached to an F, H:+1

Sam Wentzel 1F 14B

n is actually number of moles of electrons TRANSFERRED in the reaction.

Example: X 2+ + Y --> Y 2+ + X
By looking at the charges we can see the transfer of two e-, therefore n = 2 here.

Sam Wentzel 1F 14B

Write entire molecule :^)

Sam Wentzel 1F 14B

Anode is initially negative (like an anion) and becomes positive as it gives e-'s to the cathode.
The Cathode is initially positive (like a cation) and becomes negative as it receive's e-'s from the anode.

Sam Wentzel 1F 14B

Isobaric means no energy flow. Usually use it for calculating delta G.

Sam Wentzel 1F 14B

Assume 25* C

Sam Wentzel 1F 14B

It means that the path you take to get the final value does not matter. You can sum final - initial.

Sam Wentzel 1F 14B

Reversible has overall entropy increase of 0. Irreversible has entropy increase of positive.

Sam Wentzel 1F 14B

No, use .08206

Sam Wentzel 1F 14B

Porterhouse steak with mashed potatoes

Sam Wentzel 1F 14B

Oxidization agent oxidizes the reducing agent, which has an oxidation number that gets reduced.

Sam Wentzel 1F 14B

I've seen frequently in various homework problems that the solution manual solves using the expression delta H = q. I was wondering if this is always true, or if there are specific conditions for which this is true.

Sam Wentzel 1F 14B

Decide based on whether or not pressure is constant. (delta p = 0 or not)

Sam Wentzel 1F 14B

8.314 for every R outside of PV = nRT

Sam Wentzel 1F 14B

I just try to do my best, and not worry about the outcome after.

Sam Wentzel 1F 14B

No they can't in reference to this particular class.

Sam Wentzel 1F 14B

n means number of mols, in unit mols

Sam Wentzel 1F 14B

Isothermal means constant T (delta T = 0) adiabatic means delta q = 0, no change in heat ransfer

Sam Wentzel 1F 14B

You don't need to memorize them because they are provided. If they aren't the ones on the sheet, they will be given.

Sam Wentzel 1F 14B

Usually assume system is closed yes. Isolated or Open is unusually and should be specified.

Sam Wentzel 1F 14B

Temp = degrees K or degrees C. Relationship is 1 to 1

Sam Wentzel 1F 14B

Ask your TA via email for exam review.

Sam Wentzel 1F 14B

Kingdom

Sam Wentzel 1F 14B

q should be negative if the system being referenced loses heat during a heat transfer, and q should be positive if the system being referenced gains heat during the heat transfer.

Sam Wentzel 1F 14B

Delta H = 358.8 KJ means that to produce 4 mol CS2, we need 358.8 kJ (per 4 mol CS2)

371 kj / (358.8 kJ) * 4 mol CS2 = mols CS2 produced

Hope this helps!

Sam Wentzel 1F 14B

First to find q

q = 0.201mol * 3/2 * R * 11.1K

Then, because delta v = 0, and w = -P*delta V, w = 0

delta U = q + w therefore delta U = q

hope this helps!

Sam Wentzel 1F 14B

Degeneracy would only be equal to one if there were only one possible state, yes (because the expression would be 1^n, which always equals 1).

Sam Wentzel 1F 14B

No, I don't think they can be because you can't make a substance lose heat/energy in order to raise its temperature

Sam Wentzel 1F 14B

q = -w if system is DOING work

q = +w if system is HAVING work done on it

Sam Wentzel 1F 14B

In that case, we only use R = 8.1345 for work calculations right? and R = 0.08206 for PV = nRT correct?

Sam Wentzel 1F 14B

A sample of an ideal gas in a cylinder of volume 4.31 L at 298 K and 2.55 atm expands to 8.47 L by two different pathways. Path A is an isothermal, reversible expansion. Path B has two steps. In the first step, the gas is cooled at constant volume to 1.22 atm . In the second step, the gas is heated ...

Sam Wentzel 1F 14B

Say pKA = 2

To go to kA, do 10^(-2)

Hope this helps.

Sam Wentzel 1F 14B

If not indicated you can usually assume the system is under standard conditions (25* C).

Sam Wentzel 1F 14B

He's talking about the breaking of the INTERmolecular forces between water molecules. A weak force of attraction forms between the partially negative oxygen of one water molecular, with the partially positive hydrogen of another molecule, forming a slightly stronger bond called a hydrogen bond. This...

Sam Wentzel 1F 14B

Use the inverse for the reverse reaction!

Sam Wentzel 1F 14B

PKW is a constant that relates the forward and reverse reactions of acids and bases. Just think of it as a constant that facilitates finding either pKa or pKb given either one.

Sam Wentzel 1F 14B

I believe closed system's volume can be changed through compression if the substances are gases. For solids and liquids, compression barely effects their volume so they are considered to have constant volume.

Sam Wentzel 1F 14B

If it'd a weak acid, it will only partially protonate / ionize in the system. If it's a strong acid, it will completely protonate / ionize in the system.

This determines whether or not we have to use an ICE table.

The same logic applies for weak and strong bases as well.

Sam Wentzel 1F 14B

Okay, so for the substance on the reactant size, lets call this A, it will lose some value x of its concentration in order to form products, lets call these B and C. Let's say A's initial concentration is 0.1. Because x concentration of A is used, then B and C's concentrations will increase from an ...

Sam Wentzel 1F 14B

This statement is ALWAYS true of weak monoprotic acids / bases but you can make assumptions about the x value on the bottom of the expression if the KA is 10^-3 or less, such that x on the bottom roughly = 0, but x on the top STILL = x.

Sam Wentzel 1F 14B

PH = - log [H3O+] or -log[H+]
POH = -log[OH-]
14 = PH+POH (you can use this if you've found the PH of an Acid, or POH of a base, but not vice versa).

Sam Wentzel 1F 14B

State function means that you basically only need to consider the final and initial values of that function when calculating changes. Think of it as a "state" function because you only really need to consider the initial and final "states" of the function. Change = final value - ...

Sam Wentzel 1F 14B

Percent ionization kinda means what the name says. When an acid protonates a base, the acid loses an H+, and the base gains an H+. The percent ionization is equal to the [concentration of correlating ion] / [initial concentration of original substance].

Sam Wentzel 1F 14B

It helps to look at whether the reactant is processing reactant to the products side, or product to the reactant side.

Ex:

A + B --> 2C

A & B decreases by -X going from left to right

C increases by +2x

Sam Wentzel 1F 14B

It makes sense as the relation Ka x Kb = Kw = 10^-14 shows that, if Ka decreases, Kb must increase in order for this equation to still equal 10^-14, and vice versa.

Sam Wentzel 1F 14B

It depends whether reaction is endothermic or exothermic. If reaction is endothermic, and Temperature increases, then Kc will increase. If Temperature decreases, then Kc will decrease. If reaction is exothermic, and Temperature increases, then Kc will decrease. If temperature decrease, Kc will incre...

Sam Wentzel 1F 14B

Use the ice chart whenever you can't we can't determine either Kc, or final concentration of a species otherwise. Usually a value or two will be "missing" and only by using an x chart can you find them.

Sam Wentzel 1F 14B

P = Pressure, usually given in Bars / atm
V = Volume, in Liters
n = number of mols
R = The gas constant, equal to 8.3145 J mol^-1 K^-1
T = Temperature in Kelvin

Sam Wentzel 1F 14B

He announced this last quarter I believe, via email.

Sam Wentzel 1F 14B

I like to do all of the listed textbook problems a few days before the test, all for the unit that the test covers. This way, I can determine what I haven't already gathered from the lectures, and fill in the gaps by studying from the textbook. Hope this helps!

Sam Wentzel 1F 14B

Convert to ATM, so that the units cancel when plugging into the equation. Hope this helps!

Sam Wentzel 1F 14B

Kelvin. You need to make sure the units cancel when doing these problems.

Sam Wentzel 1F 14B

It depends on the state of the materials given. If the chemical equation involves gases, usually partial pressures. If the chemical equation involves AQ state chemicals, then usually use concentrations.

Sam Wentzel 1F 14B

Equilibrium Constant (Kc or Qc) = concentration of products^(stoichiometric coefficients) / concentration of reactants^(stoichiometric coefficients)

Hope this helps!

Sam Wentzel 1F 14B

Knowing the ranges is fairly important, but usually the jump from principle quantum number is noted in the problem.

Sam Wentzel 1F 14B

Both values correlate to acid strength. The lower the KA / pKA the stronger the acid.

Sam Wentzel 1F 14B

The only real exception to standard units is kg for mass, so as to cancel Joules properly (kg^2 m^2 / s^2).

Sam Wentzel 1F 14B

Chlorine has a higher electronegativity, so it pulls H's electron (within the O-H bond) closer to the C, making the H more "available" to be donated, resulting in higher acidity.

Sam Wentzel 1F 14B

Water can gain an H+ to form hydrogen peroxide (acting as a base), H3O+. It can donate an H+ to another substance, forming OH-, and acting as an acid.

Sam Wentzel 1F 14B

A salt is an ionic compound.

Cation + Anion = salt

Na+ + Cl- = NaCl :)

Sam Wentzel 1F 14B

I was a bit strapped for time near the end and did have to guess on a few.

Sam Wentzel 1F 14B

I believe it does actually. However, it is not considered a COMMON structure for coordinate compounds with a coordination number of 6.

Sam Wentzel 1F 14B

I use quizlet to memorize the shapes and bonding angles, and whether or not they are considered symmetrical (dipoles cancel) to determine if they are polar.

Hope this helps!

Sam Wentzel 1F 14B

[OH-] = concentration OH [H+] = concentration H+
pH = -log[H+] pOH = -log[OH-]

14 = pH+pOH

[H] = 10^(pH)
[OH-] = 10^(pOH)

Hope this helps!

Sam Wentzel 1F 14B

I believe shapes whose names denote "planar", such as trigonal planar or square planar, indicate a planar structure rather than a 3D structure. So planar structures are effectively considered to be lying on a flat plane.

Sam Wentzel 1F 14B

For coordinate compounds, the metal will always be at the end of the structure within the brackets, but not necessarily listed at the end of the entire molecule.

Sam Wentzel 1F 14B

For Coordinating memorize, will we need to memorize the denticities, donor atoms, ligand names etc for each ligand?

Essentially, what will we need to memorize pertaining to this subject, and what will we be given?

Sam Wentzel 1F 14B

Yes. You should memorize which molecular shapes are symmetrical, and thus non-polar assuming that the dipoles cancel, and which one's are not. Symmetrically shaped molecules will tend to be nonpolar, asymmetrically shaped molecules will tend to be polar. Hope this helps!

Sam Wentzel 1F 14B

Yes. Sig Figs count all numbers proceeding and preceding the decimal point, unless the only number before the decimal point is a 0, and 0's preceding a non-zero after a decimal point a not counted, only numbers following and including the non zero are counted. Ex: 0.00305 there are 3 sig figs, being...

Sam Wentzel 1F 14B

Oxidation state essentially means oxidation number, and it generally corresponds to periodic trends in terms of predicted charges. For example, the periodic table charge trend states that the predicted charge of an Na ion will be 1+. +1 in turn will also be the oxidation state of Sodium as well. As ...

Sam Wentzel 1F 14B

My understanding of hybrid orbitals is that they are a result of the sharing of electrons between two atoms. Due to this sharing, these electrons kind of "blend" properties of the orbitals of their parent atoms. Thus, we are left with an average of the various kinds of orbitals that the pa...

Sam Wentzel 1F 14B

It basically means that the symmetrical shape of the molecule, and the molecule's atoms' forces of attraction canceling out due to pulling electrons from the central atom in opposite directions both work to negate the polar ("asymmetrical") sharing of electrons that would otherwise be typi...

Sam Wentzel 1F 14B

A pi bond is considered delocalized only when the molecule has resonance structures (differing numbers of bonds around a central atom).

Sam Wentzel 1F 14B

I like to think of it as, the hybridization corresponds to the number of "things" surrounding the central atom. In this case the central atom is P, and the outside atoms are Br. There are 5 bonding pairs around P (Aka there are five "things" around P) therefore the hybridization ...

Sam Wentzel 1F 14B

Kind of? Take CH4, for example. The dipoles of this molecule will cancel, forming a nonpolar molecule. But if this molecule comes into contact with a polar molecule, such as HF, the partially positive hydrogen and partially negative flourine can induce a dipole onto the CH4 atom (aka e- e- repulsion...

Sam Wentzel 1F 14B

Final will cover everything we learned this quarter.

Sam Wentzel 1F 14B

Electronegativity decreases going vertically down a group, and going from left to right across a period.

Sam Wentzel 1F 14B

LDFs are always occurring, as electrons in an atom are never stationary. The movement of electrons creates regions of differing electron density, resulting in a higher electron density field of partial negativity, and a lower electron density field of partial positivity. These partial charges induce...

Sam Wentzel 1F 14B

This question is based off of resonance I believe. As a general rule, put the carbon atoms near the center of the molecule. In biological systems, carbon is usually the "spine" of a model. Put two hydrogen atoms at the left and right edges of the molecule. Try to make the bond numbers such...

Sam Wentzel 1F 14B

You do have to memorize them. For AP Chem I memorized them using quizlet.

Hope this helps!

Sam Wentzel 1F 14B

2A.17 Predict the number of valence electrons present for each of the following ions: (a) Mn4+ (b) Rh3+ (c) Co 3+ (d) P3+ For a), I determined the electron configuration of a neutral Mn atom, then subtracted 4 electrons from the outermost shell, leaving the Mn with a configuration of 1s2 2s2 2p6 3s2...

Sam Wentzel 1F 14B

c will always be equal to the speed of light. The speed of light does not change in a vacuum, unless acted upon by gravity. In the c = lambda nu equation, we assume we are in a vacuum with negligible gravitational force acting on the light.

Hope this helps!

Sam Wentzel 1F 14B

The first rules you need to know before doing electron configuration are: Pauli's Exclusion Principle: No more than 2 e- per orbital (aka no more than 2 e- per box, when drawing orbital diagrams). These e- must have opposite spins (one must have up spin, the other must have down spin) Hund's Rule: D...

Sam Wentzel 1F 14B

Which subshell, s, p, d, or f, is the most effective at shielding depends primarily on the shape of the orbital itself. The s orbital will be the best at shielding other electrons because the shape of the s orbital is a sphere, unlike the p orbital for example, which does not offer complete coverage...

Sam Wentzel 1F 14B

Polar molecules have uneven charge distribution, caused by differing electronegativity in the molecule's elements, or by asymmetry in molecule shape. Polar molecules have regions of partial positive charge, and partial negative charge. Non-polar molecules' atoms are either of the same electronegativ...

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