CHEMISTRY COMMUNITY

Indy Bui 1l

I do believe increasing T also increases k reaction constant.

Indy Bui 1l

I don't think we need to print anything out, it should all be online. Having your notes and scrap paper for work would be helpful though!

Indy Bui 1l

When delta G is negative, there is energy free to do work. Cell potential wants to go down. IF cell potential goes down there is a release of energy free to do work. I don’t think this is a correct. In the equation delta G= -nFEcell, if cell potential is higher, and more positive, then Gibbs free e...

Indy Bui 1l

They should give you this information in the question or somewhere on the test. It would definitely be tedious if we had to look these up by ourselves.

Indy Bui 1l

Thanks! For oxidation reduction I also usually use Leo the lion goes ger. LEO --> lose electrons Oxidation, GER--- gain electrons reduction.

Indy Bui 1l

At the same time, he said the test should be relatively straightforward and open notes. So, maybe we will be okay? Either way, good luck everybody!

Indy Bui 1l

It is timed, from 11:15am and ends 2:45pm, but we have 3 and a half hours. Hopefully that will be enough. Good Luck!

Indy Bui 1l

Yea, it can be used for any reaction order!

Indy Bui 1l

Not sure if this helps but here is the link to some of the answer keys for test 2.

viewtopic.php?f=160&t=62122&p=237630&hilit=test+2&sid=398e06555d588269350912837d954689#p237630

Indy Bui 1l

Okay, thank you I will definitely check them out.

Indy Bui 1l

If they are equal then the reaction is at equilibrium. If Q>K then the forward reaction or products are favored. The opposite is true for Q<K the reactants are favored.

Indy Bui 1l

It seems like this is just an online test with no scanning, so It seems like it would be hard to show your work and have that be graded. But maybe you will have to put in answers for intermediate steps and that could be counted as work? Im not sure, but Lavelle said the test would be straightforward...

Indy Bui 1l

This helped me, hope it helps you too. https://users.stlcc.edu/gkrishnan/rateunits.htm

Indy Bui 1l

Hey does anyone have any good online videos that go over Rates and Rate laws? Thank you! and good luck tomorrow everyone.

Indy Bui 1l

The derivation is:
deltaU = w + q
w= deltaH
q=-P*deltaV
so,
deltaU=deltaH-P*deltaV
which you can rearrange as:
H=U+PV

Indy Bui 1l

If you write out the units for all the constants in the equation you can see how they cancel out. I believe the final units should simplify out to V. I think you always use 8.314 for R because the J cancels out.

Indy Bui 1l

I believe so, also in a reaction I think it can sometimes be represented on the arrow. It isn't necessarily the opposite of a intermediate, an intermediate is created and consumed. The catalyst is just a substance or molecule that a reaction takes place on. It isn't used up in the reaction.

Indy Bui 1l

I could be wrong but I remember it being on the equation sheet for test 2? but otherwise it isn't on the current online equation sheet. Although the test is open notes so just make a note of it.

Indy Bui 1l

Oh wow thank you! This video was pretty helpful for me

Indy Bui 1l

I think you can convert to atm. Cause equilibrium can also be calculated through pressure. Someone correct me if I am wrong

Indy Bui 1l

You are correct, the negative is used to show that the reactants are being used up.

Indy Bui 1l

Just like qV Is heat at a constant volume, qP is heat at a constant pressure. Good luck!

Indy Bui 1l

Usually it should tell you if it isn't taken at equilibrium. This really isn't a big problem, just make sure you put the H+ concentration in the correct place in the ice tables.

Indy Bui 1l

Conjugate seesaw has to do with conjugate acids and bases. A weak acid will have a strong conjugate base. A strong acid will have a weak conjugate base. You can see how this is metaphorically related to the idea of a seesaw.

Indy Bui 1l

Intermediates are what they sound like. They are molecules that are produced in one reaction and consumed by a second reaction. They occur, because most reactions don't always happen in a single step. Intermediates just represent products in a first reaction and reactants in a second. You can tell, ...

Indy Bui 1l

You shouldn't really need to manipulate the equation too much. It is used to solve for K values for a certain recreation at different temperatures. So If you know the K at a certain temperature, you can use the equation to solve for K at another temperature.

Indy Bui 1l

I would know it since it was on the outline. It is a fairly simple concept, watch some videos on it.

Indy Bui 1l

You don't have to flip the values in the table. You can leave them and just plug them straight into the Cathode-Anode Equation.

Indy Bui 1l

Platinum is also very oxidation stable in comparison to most metals.

Indy Bui 1l

It would obviously be more helpful for the final to know things more in-depth, but for the most part what Lavelle has said during lecture should be sufficient in my opinion. If you have extra time, I would go ahead and look over the textbook though. Lavelle wouldn't put anything in outside of our ca...

Indy Bui 1l

I usually balance the Hydrogens based off Oxygens, so I guess you could say balance O2 first then H+

Indy Bui 1l

Generally no, H2O is probably best to be used. OH can create issues with PH if it is an acidic or basic reaction

Indy Bui 1l

Extensive represents things that are dependent on amount. While intensive represents things that aren't dependent on amount, like density. It is important to consider these when using constants and equations during your calculations.

Indy Bui 1l

I think this is because we want to find one mole's worth, which is avagodros number in terms of atoms or molecules.

Indy Bui 1l

it is e to the power, to get rid of Ln. Try some examples on your calculator.

Indy Bui 1l

I believe it has to do with the relationship between E and delta G. When E is positive, delta G should be negative meaning it is spontaneous.

Indy Bui 1l

I think it is because delta S is negative. So in the equation DeltaG=DeltaH-TDeltaS, delta S being negative would make delta G positive, increasing T would just make it more positive, because delta G is positive when it is non-spontaneous. (-T*-deltaS is going to equal a positive number)

Indy Bui 1l

There has to be a solid metal present in a cell diagram, Pt just happens to be the most common that is used if one isn't presented in the question.

Indy Bui 1l

Im not exactly sure if this is what you are asking but I hope this answer helps. A reaction is exothermic if Kc decreases with an increase in temperature, as the reaction shifts left, or Kc increases with a decrease in temperature, as the reaction shifts right. This is because in an exothermic react...

Indy Bui 1l

No it is not necessary. Just make sure you plug in the correct values for R, F, and T. For R you should usually be using the 8.314J/K*mol.

Indy Bui 1l

A catalyst is not used up in the reactions. An intermediate is created and later consumed.

Indy Bui 1l

Larger molecules have more entropy, due to them containing more bonds and therefore more ways of orientation. I believe that standard entropy increases with increasing mass

Indy Bui 1l

Pt is included if their is not a metallic solid in the cell diagram. Hg(l) mercury can also be included but I believe 99% of the time it is a metal solid that is included.

Indy Bui 1l

Pt is included if their is not a metallic solid in the cell diagram. Hg(l) mercury can also be included but I believe 99% of the time it is a metal solid that is included.

Indy Bui 1l

Increasing temperature flows away from "heat", so in an endothermic reaction it flows to the products and increases the equilibrium constant

Indy Bui 1l

The R constant is dependent on the units of pressure you are given. Looking up a table is helpful and just pay attention to the units you are given in the question. You don't need to memorize this, just know which values correspond to each unit of pressure.

Indy Bui 1l

(en) refers to ethylenediamine
here's its formula: NH2CH2CH2NH2
it is a chelating ligand (bidentate)

Im not sure what (os) but if you meant (ox) then ox is oxalate: C2O4 with a 2- charge, this is also bindendate chelating
Os could also mean Osmium? Where exactly did you see (os)

Indy Bui 1l

You should know the equation and the idea behind acid rain. Carbon dioxide reacts with water in The atmosphere to form carbonic acid, which is released in acid rain H2O + CO2 --> H2CO3 it may also be helpful to know that H2CO3 can dissociate and give off one of the protons H2CO3 --> HCO3- + H+

Indy Bui 1l

Yeah it would be six. I think I was a bit confused earlier when I was typing my answer. Sorry!

Indy Bui 1l

It is just how many times a ligand attaches to the metal in the center.

Example: If it attaches four times its tetradentate

Indy Bui 1l

The conjugate base of an acid is the form the acid takes after giving up a proton (H). Additionally, a conjugate acid of a base is the form a base takes after gaining a proton (H). for example: CH3CO2H is an Acid and CH3CO2- is its conjugate base, because it lost a proton H2O becoming H3O+ (hydroniu...

Indy Bui 1l

It is Hexadentate.

Indy Bui 1l

Yea I don't believe we need to know about bond order.

Indy Bui 1l

I think it is because the non-doublebonded Oyxgens have a -1 formal charge and the double bonded oxygens have a formal charge of 0. The -1 formal charge allows oxygen to want to bind. I also think that the double bonded oxygens are often on the other side of the bonded oxygens this could contribute ...

Indy Bui 1l

Boron is relatively electropositive and especially when it is bounded to fluorine, the negative charge is further distributed away from Boron. This allows it to want to accept an electron pair, making it a Lewis acid.

Indy Bui 1l

I don't think it would necessarily be raised, but if a molecule chelates then less of that molecule is needed to bond to the metal. So there would just be less ligands needed in the coordinating complex. The coordinate number doesn't increase but there are less ligands, so the charge on the metal se...

Indy Bui 1l

What is an Organic acid and what do we need to know about it?

Indy Bui 1l

No I don't believe we need to know anything about equilibrium tables in terms of calculations but I think we need a simple grasp on the concept of equilibrium as a whole. Since he didn't cover it in lecture it probably won't be on the final.

Indy Bui 1l

The molecule is polar just because of the hydrogen bonded to one of the Oxygens. This makes the entire molecule not symmetrical, making it polar, even though the shape of the central nitrogen is trigonal planar.

Indy Bui 1l

I believe the question says it is reduced TO 12% of its total, not reduced by 12%.

You would use -Log[0.12] and compared to Log(1), which is 0 the difference is just the value of -Log[0.12]

Indy Bui 1l

pH=-Log[Hydronium ion concentration]
pOH=-Log[Hydroxide concentration]

pH+pOH=14, so
14-pOH=pH

You can convert all of these into pH and then rank them.

Indy Bui 1l

The two chlorines in cisplatin get replaced by Water which bonds to the nitrogen lone pairs on two guanines. I believe these are hydrogen bonds with the Oxygen facing towards the Platinum

Indy Bui 1l

What is the biological significance of B12?

Indy Bui 1l

Elizabeth Johnson 1I wrote:
Giselle Littleton 3K wrote:Inorganic compounds are more likely to completely deprotonate.

can someone define what deprotonate means?

Deprotonate is essentially just losing a proton, you can also think of this as losing a hydrogen+. (hydrogen has one proton)

Indy Bui 1l

Yeah its probably too late, sorry. But you can still do well in the class if you do well on the final and continue to do the homework and chem community!

Indy Bui 1l

You should think of this molecule in the 3d form. In a tetrahedral, no matter where you put the atoms the Chlorine atoms will always be next to each other. I can't Draw a picture on here, but just look one up it might be helpful. The lewis structure looks like u can orient the chlorines to be opposi...

Indy Bui 1l

The smaller the cation the higher Polarizing power. Respectively, the larger the anion the higher its polarizability.

Molecules with more polarizability tend to have a more covalent character.

Indy Bui 1l

You have to memorize them, don't rely on a table on the test. I believe you need to know bond angles, shape, and whether molecules are polar or not.

Indy Bui 1l

All atoms have the potential to form temporary dipoles, due to electrons randomly moving into favorable positions. These temporary dipoles are London Forces.

Indy Bui 1l

The lone pairs cannot be on opposite sides of each other, this is due to the nature of tetrahedral shapes. if you put two lone pairs into a tetrahedral they will always be next to each other, not on opposite sides. Its hard to draw a tetrahedral model on here but looking one up might be helpful.

Indy Bui 1l

The best way to do this is to draw the lewis structures of each molecule. Once you draw the lewis structure and determine the shape you can figure out if its polar or not. Another option is drawing and canceling dipole moments on the lewis structures.

Indy Bui 1l

Can someone explain why a a single bond is a sigma and why a double bond has a sigma and a pi bond, and the same for triple bonds? thanks

Indy Bui 1l

Yes, a triple bond consists of two pi bonds and one sigma bond.

Indy Bui 1l

It will probably focus mainly on stuff after the midterm but that doesn't mean you can forget everything else we have learned.

Indy Bui 1l

Iodine has a larger atomic radius and more electrons than the other halogens. This makes it stronger.

Indy Bui 1l

Is there any easy way to remember bond angles as they correspond to shape? or do I just have to memorize them as they are.Thanks!

Indy Bui 1l

Awesome thank you for the responses!

Indy Bui 1l

Molarity is Moles/Liters. You can think about it as an amount (moles) that is diluted by a liquid (liters). So in a dilution, only the amount of liters changes and the moles stay constant and you can use the equation m1v1=m2v2. (m being moles and v being volume in liters) . Just be familiar with the...

Indy Bui 1l

Can someone please explain Dino Nuggets 11.b) Here is the question for reference "If a hydrogen electron goes from n=6 to n=4, will ∆E be negative or positive? Will the energy of the photon emitted be negative or positive?" I don't quite understand how to determine if the energy of the pho...

Indy Bui 1l

Im a bit confused, so do small cations cause greater polarizing power or do larger cations?

Indy Bui 1l

Second ionization energy is higher, because with one less electron there is greater pull from the nucleus on the remaining electron, and less electron-electron repulsion. This greater pull makes it much harder to remove the second electron.

Indy Bui 1l

The second ionization energy is higher for all elements

Indy Bui 1l

Dr. Lavelle mentioned during lecture that there was a current attached to the detector. I believe the detector was positively charged, possibly attracting the electron and detecting it, without it having a true Kinetic Energy. Although theoretically, it is emitted with 0KE, im not sure that it works...

Indy Bui 1l

Ionization energy (the energy required to remove an electron) is typically lower for s-block metals. Search up an image of ionization energy in relation to the period table for a more visual answer. But generally, ionization energy gets higher as you move from left to right of the periodic table and...

Indy Bui 1l

viewtopic.php?t=5742

Dr. Lavelle answered a similar question here ^^

Hope this helps

Indy Bui 1l

Technically you can label the axes in any order you want, however the general rule is: x, y, z Hund's rule states, however, electrons must fill one in each P axes before they are paired. This means that there has to be one electron in Px, Py, and Pz, before there are paired electrons in Px, for exam...

Indy Bui 1l

You have to refer to Hund's rule. Each orbital (for example Px,Py,Pz) must be filled with one electron before you pair electrons. Take the element Nitrogen for example. Nitrogen's orbital notation at its ground state would be written as 1s1 2s2 2p3 In the P state, nitrogen has three unpaired electro...

Indy Bui 1l

Threshold frequency refers to the minimum possible frequency of light needed to eject an electron from a metal's surface. Work function refers to the minimum energy level needed to eject an electron from a metal's surface. Thats the main difference; threshold frequency refers to the minimum frequenc...

Indy Bui 1l

These line groups represent changes in energy levels for electrons. When electrons drop an energy level (n=2 to n=1) then energy is emitted. Depending on which energy level the electron is dropping from and to, the atom emits different frequencies of lights. Experimenters measured these wavelengths ...

Indy Bui 1l

Yes, electrons generally interact with only One proton. Electrons are only ejected if the protons have enough energy, this is gained through increasing the frequency of light, not just the intensity. Once the threshold is passed, increasing intensity produces more photons and more photons interact w...

Indy Bui 1l

I believe this concept refers to the particle properties of light. If light was just a wave then increasing the intensity of light should eject electrons, but that wasn't what happened in experiments. Energy has to be high enough per photon to induce electron removal from a metal surface. Increasing...

Indy Bui 1l

I believe N subscript A refers to avagodro's number.

Indy Bui 1l

Planck's constant is used to relate the relationship between frequency and the energy of a proton, so it is used in the equation E=hv, h being Planck's constant.

Indy Bui 1l

An easy way for me to remember the 7 diatomic elements is: HOFBrINCl - H2,O2,F2,Br2,I2,N2,Cl2. Hydrogen, oxygen, fluorine, bromine, Iodine, Nitrogen, Chlorine
it sounds like Hoffbrinkle, hope this helps!

Indy Bui 1l

No, it is not mathematically different than molecules or atoms. Use avagodro's number and it won't impact your calculations.

Indy Bui 1l

It depends on the question. If you are using a formula that requires kg then of course convert to kg for the calculations or vice versa. In terms of your answer, I don't believe you will be marked incorrect for either form. Use proper judgement, if the question asks for your answer in grams then do ...

Indy Bui 1l

Combustion is essentially the reaction between a hydrocarbon or organic compound (C3H4 for example) and oxygen gas (O2), this usually occurs through burning. The products that are formed are always Water (H2O) and Carbon Dioxide (CO2). Here is a simple example combustion reaction with methane: Unbal...

Indy Bui 1l

Leading zeroes don't count. For example, "0.001" or "002" each only have one significant figure. Any zeroes coming before a number do not count towards sig figs. Sandwiched zeroes do, however, count "1001" has four sig figs and "202" has three. zeroes coming a...

Indy Bui 1l

You need to solve parts a) and b) in order to solve part c. The mass of AlCl2 you calculated in part B) is used as the theoretical yield. The question gives the actual yield as 300g. All you need to do is solve the equation 100% * (Actual Yield)/(Theoretical Yield). For my answer I got 100%* ((300g)...

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