CHEMISTRY COMMUNITY

BCaballero_4F

The chamber itself is releasing heat, not absorbing, giving a -q

BCaballero_4F

kr is the rate constant of the reaction, while k' is the rate of the reverse reaction, their relation can be shown as kr/k'=K

BCaballero_4F

From my understanding in lecture, the catalyst in this reaction is NO, not O2

BCaballero_4F

CO is zero order and because it is not in the slow step e assume its concentration does not have an affect overall

BCaballero_4F

Mols per liter per time unit which is usually in seconds

BCaballero_4F

I believe it is omitted because it is in its most stable state

BCaballero_4F

You would include whatever has (g) and (aq) from the reaction, and not solids or liquids

BCaballero_4F

Equilibrium concentration should be in Molarity which is mols/liter

BCaballero_4F

Yes because (g) and (aq) are included in the formula for K

BCaballero_4F

Yes you would in order to find the equilibrium concentrations

BCaballero_4F

You add it on the side with the higher charge, and you add based on however much is needed for it to be balanced.

BCaballero_4F

ASetlur_1G wrote:Is it correct to say that G(not) is used at equilibrium (K value) and G is used for other conditions (Q value)?

Yes this is essentially correct to say

BCaballero_4F

n refers to the moles of electrons so the coefficient in front of the electrons when you balance the half reactions

BCaballero_4F

In the diagram the anode is on the left so that is where the oxidation should be

BCaballero_4F

In the anode is where oxidation occurs (so electrons are lost) and the cathode is where reduction occurs (electrons are gained here) and the flow is from anode to cathode.

BCaballero_4F

You would need to use the delta H equation and work with what you are given in the problem, mainly find what is constant and manipulate the equation to reflect that while looking for delta H.

BCaballero_4F

PranaviKolla2B wrote:What is the difference between a calorimeter and a bomb calorimeter?

I believe a calorimeter is just a thermally insulated container while a bomb calorimeter is a sealed container that is metal in water which is also in an insulated container.

BCaballero_4F

Oxidation means losing electrons, while the oxidizing agent is doing the oxidizing to the other species, so it is then being reduced.

BCaballero_4F

The reducing agent does the reducing, so in this process it is oxidized. The oxidizing agent does the oxidizing, so it is reduced.

BCaballero_4F

It is very helpful to split into half reactions when we are trying to balance them, so I think that is why

BCaballero_4F

If delta G is negative then it is spontaneous, and if it is positive then it is not spontaneous.

BCaballero_4F

I dont think he will curve the midterm considering the final grade is technically curved

BCaballero_4F

Yes, this applied to elements in their standard state

BCaballero_4F

To calculate this you would add together the bonds broken, then subtract that number by bonds formed.

BCaballero_4F

You can look at adding heat as adding a product or reactant depending on whether it is exothermic or endothermic. If you are adding a product, then it will shift to reactants, and vice versa.

BCaballero_4F

This refers to q.

BCaballero_4F

You basically alter the reactions to get the reaction you want, by multiplying or moving reactants to products and products to reactants

BCaballero_4F

The 1st law of thermodynamics is conservation of energy, so the energy of an isolated system must remain constant. This gives delta u = q + w = 0

BCaballero_4F

Delta H would be positive since breaking a bond would be considered an endothermic reaction

BCaballero_4F

Whether or not delta H is positive or negative depends on the reaction (if it is endothermic or exothermic)

BCaballero_4F

You use an ICE table when you are looking for the concentration of something. Ka helps to solve for x

BCaballero_4F

Yes, because it is a gas and we include gases in Kc. (just depends on state given)

BCaballero_4F

You go back and use the quadratic formula only if it is more than 5%

BCaballero_4F

It would be a strong acid instead of a weak acid

BCaballero_4F

Moles and liters count for sig figs in this question

BCaballero_4F

Which one you would use depends on the units you are initially given in the problem

BCaballero_4F

Since it is H2O as a gas, you would still include it.

BCaballero_4F

I believe it is anything less than 10^-3

BCaballero_4F

You use the stoichiometric coefficients

BCaballero_4F

Kp is for pressure and Kc is in terms of molarity

BCaballero_4F

Autoprotolysis can be shown using H2O and is the proton transferring of two of the same atom

BCaballero_4F

Q does not have specific units because it is a ratio.

BCaballero_4F

Yes it should be okay, last quarter as long as it was recent we received credit. It is also still pretty early since it's week 2 and we went over it last week.

BCaballero_4F

Yes in this case you can ignore pressure

BCaballero_4F

Q can be calculated at any given moment, not only when the system is not at equilibrium, so Q=K is when Q is calculated when it is at equilibrium.

BCaballero_4F

Because at equilibrium the constant is consistent so if one is higher than the other would also be higher since that constant must be maintained

BCaballero_4F

This means that at equilibrium more products are present than reactants. In this case, a lot more

BCaballero_4F

K is used for when it is at equilibrium

BCaballero_4F

No, the concentration of products and reactants are not the same but the ratio that gives the equilibrium constant is the same.

BCaballero_4F

I believe the answer would be A, because the equilibrium constant gives a ratio which includes products and reactants at equilibrium. For example, when K is greater than one the reaction has mostly products and when K is less than one the reaction has mostly reactants.

BCaballero_4F

Lewis acids accept electrons and Lewis bases donate electrons

BCaballero_4F

pH measures acidity while pOH measures basicity, and you can use one to get the other

BCaballero_4F

The different forms of a molecule because there can be more than one

BCaballero_4F

Yes all ligands are basically Lewis bases

BCaballero_4F

In most cases Ammonia is considered a weak base

BCaballero_4F

Because a sigma bond is stronger and more stable so there will be one of these, and pi bonds would come after the sigma bond because they are weaker.

BCaballero_4F

The energy of the hybridized molecule is an average of the beginning two.

BCaballero_4F

To explain bonds more thoroughly

BCaballero_4F

You begin with the Lewis Structure which then allows you to use VSEPR to predict the shape, and from these two you should know the number of regions of electron density which would then allow you to derive the hybridization.

BCaballero_4F

005384106 wrote:Does hybridization depend on the number of bonds between atoms?

The number of hybrid orbitals depends on the number of regions of electron density.

BCaballero_4F

Cl is the central atom because it allows for stability

BCaballero_4F

Yes, but only sp3 + it corresponds with electron densities

BCaballero_4F

The electrons in sigma bonds are very close to the nuclei, while the electrons in a pi bond are further from the nuclei so the attraction is not as strong, thus making them easier to break.

BCaballero_4F

Yes they do count, so for this example it would be 4 instead of 3.

BCaballero_4F

I believe we will need to know how to draw them but I am not sure, regardless it would be beneficial.

BCaballero_4F

I believe the f-block can store electrons too since the octet exception applies to those after the third period.

BCaballero_4F

I believe it will be specified but if not the noble gas configuration should be sufficient.

BCaballero_4F

Usually the question will specify whether they want the whole thing written out or not.

BCaballero_4F

If electrons are shared equally, it is nonpolar. If electrons are shared unequally, it is polar, which you should be able to see once you calculate formal charge.

BCaballero_4F

The most electronegative atom should have the negative charge, I think it does matter what the formal charges are on individuals which would then lead to the correct charge for the overall structure.

BCaballero_4F

A coordinate covalent bond is between an acid and a base where one acts and the lone electron donor and the other acts as the lone electron acceptor.

BCaballero_4F

I think the midterm will be all free-response because all of the practice problems given usually are as well.

BCaballero_4F

Yes, the most stable structure will make it the most accurate so it is the one you should draw.

BCaballero_4F

You get the number of valence electrons from whatever group the element is in.

BCaballero_4F

I would recommend counting the formal charge so you have the most similar structure to what the real structure looks like.

BCaballero_4F

It may also help to understand that second row electrons can only have an octet due and count the electrons up to the 2nd row (1s^2,2s^2,2p^6) and if they are in a different row such as the third row they can take e- from the d subshell.

BCaballero_4F

You would put the atom that is the least electronegative or has the lowest ionization energy in the middle.

BCaballero_4F

It means it is moving to the next orbital because it is excited.

BCaballero_4F

If the negative is referring to the charge then yes you would add an electron for every negative shown and subtract an electron for every positive shown when referring to charge.

BCaballero_4F

Anions are bigger because an electron is added. Cations are smaller because a valence electron is lost.

BCaballero_4F

I don't think you will be penalized for moving on from Quantum World if you wanted to.

BCaballero_4F

For the most part I think it will be given unless it is simple for us to figure out.

BCaballero_4F

As long as the overall charge is accurate and the formal charge is matched then you should be fine.

BCaballero_4F

The central atom has to be the atom with the lowest ionization energy, so I think those can be in the middle if they have a lower ionization energy than other atoms involved.

BCaballero_4F

This is due to their shape

BCaballero_4F

Schrodinger's uses a function to describe an e- in an atom and uses the concept that an e- with wavelike properties and indeterminacy in momentum and position can be described by a wave function.

BCaballero_4F

The electron has to be in ground state for the radius to be measured properly, not excited.

BCaballero_4F

They are due by Sunday, Professor Lavelle explained in an email how when discussing Chemistry Community the week "ends" on Sunday.

BCaballero_4F

Light does not have mass, which is a required variable in the De Broglie equation.

BCaballero_4F

Definitely take advantage of all the tutoring they offer and practice as much as you can, YouTube also helps a lot.

BCaballero_4F

Combustion consists of burning, usually O^2 reacts with something else and CO^2 and H^2O are products.

BCaballero_4F

I find it helpful to list the elements on both sides (under reactants and products) and then plug in numbers that look correct and update the numbers as I go by crossing them out and putting the new ones.

BCaballero_4F

You would use the Rydberg equation to calculate the wavelength of light when an electron moves energy levels.

BCaballero_4F

Yes, however I think it is best to show work so that in case you make a mistake you may receive partial credit.

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