CHEMISTRY COMMUNITY

rabiasumar2E

A catalyst is first a reactant and is used up, then is a product and is formed.

rabiasumar2E

The exam starts at 11:15am and is over at 2:45pm! It will most likely close at 2:45pm and submit whatever you had completed on it.

rabiasumar2E

It will probably be given to you in the question!

rabiasumar2E

The slow step determines the rate of the full reaction because no matter how fast the other steps are, the reaction is only complete once the slow step has been completed.

rabiasumar2E

There isn't a numerical value that you would use to deduce if it's a slow step or not. It would just be the slowest step out of all of the steps.

rabiasumar2E

My TA emailed us about where to pick them up! You should get one from yours soon!

rabiasumar2E

[A]0 represents the initial concentration and [A] represents the concentration of the reactants!

rabiasumar2E

I'm pretty sure you'll get them back during your section this week !

rabiasumar2E

Yes! Rate constants are always positive because concentrations cannot be negative values.

rabiasumar2E

Adding carbon graphite isn't as common as adding platinum but it can be written wherever platinum can be written! Writing carbon graphite is equivalent to writing platinum in this circumstance.

rabiasumar2E

You add Pt to your cell diagram when there is no conducting solid in any of your half-reactions.

rabiasumar2E

Use Pt in your cell diagram if there is no solid present in the reaction!

rabiasumar2E

Yes! Galvanic cells are spontaneous and therefore, the E must be positive.

rabiasumar2E

I'm sure there are a few that we'll have to derive on our own but I think he'll give us every equation we need to do so with on the equation sheet!

rabiasumar2E

The purpose of a salt bridge is to maintain charge balance.

rabiasumar2E

Whatever is being oxidized is the reducing agent and whatever is being reduced is the oxidizing agent.

rabiasumar2E

Test two is going to cover everything from the second page of outline four and all of outline five !

rabiasumar2E

Half reactions are the reduction and oxidation reactions that make up a redox reaction.

rabiasumar2E

The salt bridge allows ion transfer between the two solutions and is used to maintain charge !

rabiasumar2E

It represents the charge of one mole of electrons.

rabiasumar2E

Since E is an intensive property, it doesn't depend on how many times the reaction occurs and therefore, will never change! H is an extensive property, which is why you can use Hess's Law.

rabiasumar2E

E doesn't change because it's an intensive property! It doesn't depend on how many times the reaction occurs.

rabiasumar2E

After taking the midterm, I'm pretty certain he'll provide us with any constants we may need for an exam.

rabiasumar2E

S= entropy
kB= boltzmann constant
ln= natural log
W= number of states^(number of atoms or molecules)

rabiasumar2E

Because standard conditions don't change !

rabiasumar2E

It's zero in an isothermal reversible reaction.

rabiasumar2E

A reaction is spontaneous when delta G is negative!

rabiasumar2E

'Rev' just means reversible so q rev is used when calculating the heat for a reversible reaction.

rabiasumar2E

The units for change in enthalpy is J/K. And no, the units for temperature will not remain! They'll end up cancelling out.

rabiasumar2E

Yes! When pressure increases, entropy decreases because volume decreases.

rabiasumar2E

The unit of the boltzmann constant is J/K.

rabiasumar2E

It represents the change in the internal energy of a system !

rabiasumar2E

The difference between the two work equations is that one includes the integral and one does not. You use the work equation without the integral when pressure is constant.

rabiasumar2E

It's because HF is actually a weak acid, meaning it doesn't dissociate completely!

rabiasumar2E

Another example could be hot coffee in a thermos.

rabiasumar2E

I think we probably will end up having to know most if not all of the thermodynamics outline for the midterm.

rabiasumar2E

I'm sure you can use kJ/K for entropy if it matches the units of everything else in the problem you are working on but the SI unit for entropy is K/J!

rabiasumar2E

If a reaction is exothermic, an increase in its temperature will cause the reaction to shift to the left and a decrease in its temperature will cause the reaction to shift to the right!

rabiasumar2E

The energy of the products being higher than the reactants is due to the products absorbing heat from their surroundings, meaning the reaction is an endothermic reaction.

rabiasumar2E

Solids and liquids don't affect the value of K, therefore we don't include them in the calculations!

rabiasumar2E

I usually read the sections he has assigned for homework on the outline that we're focusing on in lecture at that time !

rabiasumar2E

What the question contains will help determine what method you should use!

rabiasumar2E

The unit for standard enthalpy of formation is kJ/mol!

rabiasumar2E

No, as long as the products are in the numerator and the reactants are in the denominator the order of which one you write first does not matter.

rabiasumar2E

If a reaction requires heat it is endothermic. If a reaction gives off heat it is exothermic.

rabiasumar2E

Strong acids and bases dissociate completely whereas weak acids and bases do not. I would just memorize the strong acids/bases to make working out problems a little easier. There is a list of both in the textbook.

rabiasumar2E

Solids and liquids don't affect the value of K, so we leave them out.

rabiasumar2E

The answer is no change because the number of moles on both the reactant and product sides are the same.

rabiasumar2E

Aqueous solutions would be included in the equilibrium equation because these effect the value of K but solids and liquids aren't included because they don't affect K in any way.

rabiasumar2E

Because they don't change the value of K in any way, so you can disregard them while writing the K or Q expression.

rabiasumar2E

Test one starts on January 21st and test two starts on February 25th. So, whenever you have ur discussion after or on that date is when you'll have your tests.

rabiasumar2E

I'm not exactly sure but maybe a good way to gauge it would be to see how many ideal gas homework problems he's assigned? I would just be as comfortable with ideal gases as I could be to be prepared.

rabiasumar2E

K = 1 is rare because it's not common for reactants and products to have the same stability.

rabiasumar2E

4s would come after 3d!! In high school most of us learned to put the 4s before the 3d since they are in that order on the periodic table, but energy level wise the 4s would come after 3d. To put it simply, whenever you're doing electron configuration, put them in numerical order.

rabiasumar2E

I would memorize them, which seems like a hassle but honestly the more practice problems you do, the easier it gets to memorize.

rabiasumar2E

I would! It's probably the best way to ensure that you can name anything he might give us on the final.

rabiasumar2E

You always place radicals on the most electronegative atom!

rabiasumar2E

S character refers to the contribution of the s-orbital in hybridization. For example in a sp2 notation, there would be 33% s character and 66% p character. In a sp3 notation, there would be 25% s character and 75% p character.

rabiasumar2E

alex_4C wrote:Is our final on the 8th or 9th?

Our final is on the 8th at 11:30am.

rabiasumar2E

Honestly everything we've learned in class will most likely make its way on the final in one way or another. I would just look at the outlines from his website and continue to do the hw problems he's provided to try to be well prepared for it.

rabiasumar2E

In the book it mentions that when they say 'acid' they are referring to the Brønsted acid so I'm assuming that applies for lecture as well. I think if he were to mention lewis acids/bases he would say 'lewis acid' or 'lewis base'. And I would definitely know how to differentiate the two for the final!

rabiasumar2E

There is a list of strong acids in the book that I would recommend having memorized!
HBr, HCl, HI, HNO3, HClO4, HClO3, H2SO4

rabiasumar2E

On the periodic table, there is a diagonal band of amphoteric oxides closely matching the diagonal band of metalloids. You can also tell if it is able to accept and donate hydrogen ions. As for identifying them in a chemical equation, it's not necessarily important to know if it's amphoteric, just w...

rabiasumar2E

Yes! All amphoteric compounds are oxides.

rabiasumar2E

Yes water is considered both an acid and a base. Just as the above response says, amphoteric compounds have the characteristic of both. You can find amphoteric oxides on the period table in a diagonal band closely matching the diagonal band of the metalloids.

rabiasumar2E

A lewis acid is an electron pair acceptor and a lewis base is an electron pair donor. This differs from Brønsted acids and bases; Brønsted acid is a proton donor and a Brønsted base is a proton acceptor.

rabiasumar2E

Sigma bonds are stronger than pi bonds because they overlap head on and a greater overlap = a stronger bond. Pi bonds overlap from the side therefore their bond is not as strong because the overlap is not as much.

rabiasumar2E

A sigma bond is symmetrical with no nodal planes with respect to the internuclear axis but a pi bond has a single nodal plane with respect to the internuclear axis.

rabiasumar2E

Yes, so if two of the same atoms are on the same side of the central atom then it is cis- and if they are on opposite sides then it is trans-.

rabiasumar2E

It can occur when a hydrogen is bonded to a nitrogen, oxygen, or fluoride.

rabiasumar2E

Yes, a lone pair decreases the bond angles. For example, in NH3 since there is a lone pair the bond angles will be around 107 degrees rather than the 109.5 degrees that shape usually gives.

rabiasumar2E

Hydrogen bonds are the strongest types of dipole-dipole interactions. Ion-ion are the strongest type of IMF!

rabiasumar2E

In lecture he mentioned that we aren't expected to know the value of the actual angle and that we should just be able to determine that it would be less than 109.5 degrees.

rabiasumar2E

Yes, ion-dipole is stronger than dipole-dipole. The order goes: ion-ion, ion-dipole, dipole-dipole, dipole-induced dipole, induced dipole-induced dipole.

rabiasumar2E

You can look at the difference in electronegativity to figure out if a molecule is polar or look at its shape. If a molecule is symmetric it is non-polar, i.e. CH4 would be non-polar because it's symmetrical. An example of a polar molecule is NH3, which is not symmetric.

rabiasumar2E

A polar molecule occurs when the molecule is not symmetric, i.e. when one side has a positive charge and the other side has a negative charge.

rabiasumar2E

The VSEPR formula is AXnEn.
A: central atom
Xn: number of bonding electrons (electrons connected to central atom)
En: number of lone pairs

Draw the lewis structure and you'll find everything you need for the formula!

rabiasumar2E

I think test two covers everything we have learned after the first midterm.

rabiasumar2E

Any element in and after the third period can have an expanded octet due to them having more than just p-orbitals.

rabiasumar2E

Van der Waals is usually used to describe all intermolecular forces. LDF is always included in it.

rabiasumar2E

I would maybe ask your TA just to be sure. If your discussion session is in the beginning of the week, I'm sure doing problems from chemical bonds is okay.

rabiasumar2E

Ideally you want your central atom to have a formal charge of zero or as close as you can get to it. The octet rule can only be broken if that element is able to break it. Elements from period three or higher have d-orbitals, allowing them to accommodate for more electrons. So just make sure that if...

rabiasumar2E

Elements in period three or higher have d-orbitals in their valence shell, meaning that they are able to accommodate additional electrons because they have room to do so.

rabiasumar2E

In lecture he said that atoms in period 3 or higher can accommodate additional electrons because they have d-orbitals, allowing them room to do so. I think breaking the octet rule is done when trying to draw the lewis structure most accurately with the lowest energy or sometimes just to get the righ...

rabiasumar2E

When the bond is shorter, the attraction of the electrons to the nucleus becomes stronger. Therefore, a double bond (since it is shorter) will always be stronger than a single bond and a triple bond will always be stronger than a double.

rabiasumar2E

How do we know how many valence electrons an element with an expanded octet has? How are lone pairs shown in a lewis dot structure? An element will always have the same number of valence electrons whether it has an expanded octet! Cl can have an expanded octet but its number of valence electrons wi...

rabiasumar2E

For the most part, the central atom will only have an expanded octet because we are trying to get the formal charge to zero. The point is just to draw the best lewis structure so if your formal charge for the central atom isn't zero and it has an expanded octet, I would try to move somethings around...

rabiasumar2E

From seeing what Test 1 looked like, I'm pretty sure that the midterm will also be somewhat the same format so I'm assuming all free response.

rabiasumar2E

Lone pairs are just electrons that aren't bonded!

rabiasumar2E

A double is shorter than a single bond because it has more electrons and therefore, attracts to the nucleus more, making it stronger. A triple bond is also shorter than a double bond.

rabiasumar2E

We use the formal charge equation to help us find the most stable structure of a molecule.
in FC= V -(L + S/2), V= number of valence electrons, L= number of lone pair electrons, S= number of shared electrons

rabiasumar2E

I don't want to say that the textbook is wrong but I'm sure that neon's electron configuration is 1s2 2s2 2p6.

rabiasumar2E

I would say chemical bonds since that's what we'll be covering in lecture this week.

rabiasumar2E

When 4s is not filled, it has lower energy than 3d so it is filled first. But when 3d is filled it is no longer higher in energy, which is why Lavelle wants us to write 4s after 3d. So, when you remove an electron, you would remove it from the closest orbital which would be 4s.

rabiasumar2E

Say you want to find the formal charge of S in SO4^-2. You have to draw the lewis structure and from there you'll find the the numbers to plug into the formal charge formula. Number of valence electrons for S is 6. 'L' represents any lone pairs it may have and in this case it has 0. 'S' represents s...

rabiasumar2E

Yes, since we're still covering this topic in class we can do any five problems from the ones he assigned for the quantum world section.

rabiasumar2E

We use those elements because they are noble gasses and aren't reactive, meaning their configuration is complete. 'He' would be 1s2 and 'Ne' would end with 2p6. We use them to help with the configurations of other elements that are further down the periodic table mostly for convenience so we don't h...

rabiasumar2E

First I made sure that the equation was balanced. Then I did the following to find out how much of CO2 and H20 was used to make glucose and therefore figure out what the limiting reagent was. 10.0g CO2 (1mol Co2/ 44.011g CO2) (1mol C6H12O6/ 6mol CO2) (180.156g C6H12O6/ 1 mol C6H12O6) = 6.822g 20.0g ...

rabiasumar2E

First I made sure that the equation was balanced. Then I did the following to find out how much of CO2 and H20 was used to make glucose and therefore figure out what the limiting reagent was. 10.0g CO2 (1mol Co2/ 44.011g CO2) (1mol C6H12O6/ 6mol CO2) (180.156g C6H12O6/ 1 mol C6H12O6) = 6.822g 20.0g ...

rabiasumar2E

I believe we use it to help us find the allowed energy levels in a quantum system.

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