CHEMISTRY COMMUNITY

Michaela Capps 1l

My TA said we will probably have to identify them in complex molecules but not have to draw them individually.

Michaela Capps 1l

Yes you can use just OH because K is not involved in the redox reactions.

Michaela Capps 1l

Doesn't necessarily mean it is always reversible, but it will always be transferred as work because q=0.

Michaela Capps 1l

Collision theory has to collide at same time, same orientation and enough energy to overcome the activation energy.

Michaela Capps 1l

Delta H is equal to q at a constant pressure.

Michaela Capps 1l

I would assume both just complex because you'd have to understand what the functional groups are

Michaela Capps 1l

You're right--you do need delta H of fusion because heat is lost from ice to water.

Michaela Capps 1l

It is postitive because y=mx+b so the slope is positive.

Michaela Capps 1l

For the first question, in the integrated rate law, kt is positive, so that is why we use pos slope.

Michaela Capps 1l

Yes Alexander is right. Even though a reaction might be thermodynamically favorable, it still has to overcome an activation energy.

Michaela Capps 1l

Cp and Cv are usually just used for different conditions, either constant pressure or constant volume. I'm not sure when u would switch.

Michaela Capps 1l

Pre equilibrium is easy to use but is not as flexible at steady state approximations which is usually used for complex mechanisms.

Michaela Capps 1l

It is a part of the laws of thermodynamics.

Michaela Capps 1l

You have to sublimize Carbon to the gaseous state because deltaH has to have them all as gasses.

Michaela Capps 1l

OIL RIG Oxidation is losing (electrons) and Reduction is gaining (electrons)

Michaela Capps 1l

The molecule that is substituting the leaving group.

Michaela Capps 1l

In order to subtract the 1/T and 1/T' you have to have a common denominator. so you multiply both by the others denominator.

Michaela Capps 1l

The anode is more negative so it makes sense that it is receiving negative electrolytes.

Michaela Capps 1l

This is the idea that in a perfect system q(metal)+q(water)=0 so making overall q(water) negative, will be equal to q(metal).

Michaela Capps 1l

Yes it is because it is already in kJ/mol

Michaela Capps 1l

For calculating delta n (change in moles) we look at gas phase molecules in the chem equation because those are the only ones that will affect work done. In this problem, were given 6mol H20 and 6 mol CO2 in gas phase on products, totaling 12 moles, and 6 mol O2 on gas phase on the reactants (we ign...

Michaela Capps 1l

You would look at the standard cell potentials. The most positive is the cathode. Also, cathode-(anode) should be positive.

Michaela Capps 1l

Number of orientations raised to the number of molecules. This tells you all of the possible orientations the molecules can have.

Michaela Capps 1l

We didn't go over this derivation and also are not responsible for knowing it.

Michaela Capps 1l

When it is reversible, the system is at equilibrium so it =0. When deltaS >0 the reaction is spontaneous, which is a part of the second law of thermodynamics

Michaela Capps 1l

The half reactions only include the elements that are being oxidized or reduced. The net ionic equation will include all elements (and spectator ions) to balance the two sides.

Michaela Capps 1l

S is the residual entropy while delta S is the change in entropy.

Michaela Capps 1l

Cell potential only depends on concentration so electrode surface area doesn't affect it.

Michaela Capps 1l

You have to solve by melting the ice, and then by bringing the liquid to the final temperature (which is endothermic/requiring heat). Heat absorbed by ice cube = (50g/18.02g*mol )(6.01*10^3 J*mol^-1) + 50g*4.184J*C^-1 *g^-1 * (Final Temperature-0 degrees C) Heat released by liquid water = (400g)(4.1...

Michaela Capps 1l

I used a ratio so I set up 358.8kj/4 molCS2 = 415kj/x moles and then I solved for x.

Michaela Capps 1l

The amount ionized(starting-ending)/starting amount *100 would be the percentage

Michaela Capps 1l

It depends on if the reactants or products are aqueous. If they were, adding water would change the molarity and thus the reaction and how it proceeds.

Michaela Capps 1l

For my test 4 #6, There were only 3 values? A + B <---> 2C. If you are asking about the 2 on the 2C, it would come into play for the change line of the ICE table (and be 2x instead of x).

Michaela Capps 1l

^^^ Yes I also believe this to be right. The equation would specify by including the physical properties of the reactants and products in the equation. If any were aqueous then you would know to watch out for that

Michaela Capps 1l

Just to expand, the equation will shift towards the side with less moles of gas in an attempt to reduce the pressure and return to equilibrium

Michaela Capps 1l

Yes it would be the same number of SF as the K value

Michaela Capps 1l

The coefficients are superscripts for equilibrium expressions so when it said half coefficients, I put the half in the superscript of the x. If it said double all coefficients, you would have x^2

Michaela Capps 1l

The .5 in bond order indicates resonance. 1.5 bond order is going to be the average of single and double bonds in a resonance hybrid

Michaela Capps 1l

For covalent, the difference in electro negativity will determine if the bond is either polar or non polar. And yes the increasing number of bonds decreases the bond length. Higher bond order (number of bonds) indicated more attraction between electrons and the atoms will be held closer together

Michaela Capps 1l

Also means they have the same electronic structure and will have similar chemical properties

Michaela Capps 1l

I have memorized that there can be octet expansions for elements in the 3rd period and beyond but I don't think I really understand why that is? I usually remember things better when I understand everything versus just memorizing so can someone please explain? Thanks!

Michaela Capps 1l

We have mostly just looked at the hybridizations of elements with a maximum of d orbitals, but what would the hybridization for an atom with say an f orbital look like?

Michaela Capps 1l

Yes, even though you fill the 4s before the 3d orbital, you should write it in order of ascending energy, which would be 3d then 4s. This also makes it easier to see what electrons will be ionized (from the highest energy orbital)

Michaela Capps 1l

I believe that in general, not more that 16 electrons will be involved

Michaela Capps 1l

And in each double/triple bond, the first bond will be a sigma bond, and then the successive bond(s) will be pi bonds

Michaela Capps 1l

If you are talking about the Bohr frequency model (just because it includes E and results in frequency) then you are going to use the delta E/change in energy

Michaela Capps 1l

Why is it that n=1 is the lowest energy level?

Michaela Capps 1l

Okay so you were right to want to use E=h·ν but you rearrange it because v ("nu" or frequency)=c/λ so we can use E=hc/λ =(6.626·10^-34 Js)·( 2.997·10^8m/s)/ (470·10-9 m)=4.225*10^-19 J. The lamp emits 32J/s so in 2 seconds, the lamp emits (32*2=) 64J. So we use E(per mol of photons)=N(phot...

Michaela Capps 1l

The work function is given to us in kJ/mol but we want it per electron so we use Avogadro's constant. First I converted kJ to J by multiplying my 10^3 because that's what's easiest for me in terms of conversions. (150*10^3 J/mol) x (1 mol/6.02*10^23)...moles cancel out so we are left with 2.50*10^-1...

Michaela Capps 1l

In class, Lavelle said there were 3 sig figs and I was just confused on which zero was included in the 3 SF? Thanks!

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