CHEMISTRY COMMUNITY

rhettfarmer-3H

An intermediante is a value that is produced and then consumed in a reaction so essitially it canceled out and doesnt play affect on the rate of the reaction.

rhettfarmer-3H

I think Mathew gave us the cheat code for K units. So rate = k a to a power. So Zero is m/s. One is 1/s and second is 1/sm.

rhettfarmer-3H

Catalyst do not affect H because H is dependent on the formation of product and reactant and a catalyst cannot change those properties.

rhettfarmer-3H

How come second order has a positive slope unlike first and zero?

rhettfarmer-3H

If you are given the half life you can use the three half life equation to match the half rate and therefore, it will also allow you to find the order.

rhettfarmer-3H

If it lowers the EA of the forward it must lower the reverse too because think about it of the reaction profile. They share the Ea hump in a sense. So if it goes down it helps both ways of the reaction.

rhettfarmer-3H

A zero order reaction means that the rate does not dependent on the concertation of the element.

rhettfarmer-3H

We can use the second equation with the log under standard conditions. This is just a faster route to your answer at 1 atm and 298k.

rhettfarmer-3H

Yes, you flipp the sign if you flip the equation. It is important to remember thought hat C-A equation accounts for the flip though.

rhettfarmer-3H

I think that the CL2 is in its elemental state which means that it does need to be included because it does not have G. It is in its elemental state/

rhettfarmer-3H

I think the best way you can tell is by the value of gibbs free energy so if the gibbs free energy is negative it is spontaneous.

rhettfarmer-3H

This would be under the assumption of constant pressure. It is also notes that this equation can be expressed in terms of -dnRT which also gives you the work.

rhettfarmer-3H

I think the best way to think of this is through the enthaply graph. SO first determine if its Endo or exo. In this case h is postive therefore, it endo. Endo goes to higher energy for products so there is less EA for the reverse. SO we will take the forward and substract it from h.

rhettfarmer-3H

yes, the k' is both the notation for both reverse and psuedo rates. Therefore, it is important to remember in context what you are working on. Also, remember that k/k' is equal to the K of the equilibrium will help for some problems

rhettfarmer-3H

the half life is given on the outline which is pretty easy to remember.

First order is
t=.693/k

2nd is

t=1/kA intial

Zero is
t=Aintial/2k

rhettfarmer-3H

It comes from ln(.5) or ln(2) however you want to think about it. ln.5A-lnA=-kt. Which gives ln(-.5A)=-kt Therfore gives the equation .693/k=t.

rhettfarmer-3H

To begin, Half life is when the reaction reaches half the initial amount. Hence, 0.5A=-kt+A. Then becomes -.5A=-kt which more so .5A=kt. So then we get A*1/(2*k)=t. That the half life of zero reaction.

rhettfarmer-3H

I think the way we are trying to figure out this value for C. Is through a system of equations almost comparing two equation with only C. So equations 1 and 4. Values of A and B are the same which is a form of isolating C. Therefore when comparing the two A, B and k can be canceled out because both ...

rhettfarmer-3H

For pure memorization, first order and zero is -k. And 2nd and third are positive K. However, the reason for this is the integration of the two graphs as shown in the lecture. Hence, remembering the graphs for each will help understand the K and its slope.

rhettfarmer-3H

I think the best way to think of this is how you are going to cancel out what is already there . So for example rate=kA, so rate is m/s=KM so K must be 1/s for first order. So in theory use this idea that rate being M/S and the other side being kA. So as we increase order we increase the degree of A...

rhettfarmer-3H

sounds like you are doing it all right. I got my K value to be different. So maybe recalculate K. Using a given experiment. so get one rate=kAB^2. Other than that I think you are on the right track. C is always 1 because it is zero order, so just check you K again using the values given for one expe...

rhettfarmer-3H

I think you can just use the equation dG=-n(number of electron, in this case, looks like 3)FE. The value of G should be expressed in J then just solve for dG and you are all good.

rhettfarmer-3H

Water is add at the times where Oxygen are not balance. If you read the book it gives a pretty good clarification on how to do it and such. Therefore, it also is key to note that they can cancel out on opposite sides and it depends in the question if its basic or acidic.

rhettfarmer-3H

This problem requires to use of the chart with the listed values. In a sense, you are coupling the reaction with one of the following therefore there is a range where you can couple one reaction and not the other and need to find the value between.

rhettfarmer-3H

I think you need to use the chart they gave us with certain equations with the E values on them. Hence, You look up Iron 2 and copper 1 and on top of that you need to know which one is cathode and anode and then make sure you add them correctly care to sign.

rhettfarmer-3H

I think the best way to think of this is E=E(naught of the cathode)-E of the anode. These values are found through the table which is given to us with equations. SO look up Gold and aluminum values and subtract them take care to the sign.

rhettfarmer-3H

The best way to remeber this is between the relationship that Cv is Cp-r and therefore, you just memorize the chart of the CP.

rhettfarmer-3H

The sign of work is deoendent on who does work and who gets work done on them. If the system has done work then it's the normal negative. Therefore, if the system gets work done on it then work is positive.

rhettfarmer-3H

I think the only relationship between is w=-NRtln V2/v1 and that goes to s=Rnln(v2/v1). This comes from w/t and hence comes from s=q rec/t. Basic algebra allows for such creation and somewhat a relationship.

rhettfarmer-3H

I think the two-equation you should get is ds=qrev/t and ds=S surr + S sys. Then, it also states that spontaneous reactions will increase the entropy of the universe.

rhettfarmer-3H

I think he was referring to spontaneous reactions. hence, the second law says that the universe always increases disorder. However, there are some small changes in the universe that do restore order. So S can be negative.

rhettfarmer-3H

I think the best way to think of this is with the graph of water. So, looking at the curve you see that energy is on the y axis. So S to L is the first state of mater change. THen L to gas. Each takes more and more energy. S to L takes less energy than L to G because the molecular of a solid is comp...

rhettfarmer-3H

State property is basically just referred to the property that the intermediate values don't matter, the only thing is how thing went from initial to final. Therefore, you just need the initial and final values to calculate the answer or value of the answer.

rhettfarmer-3H

A phase change just refers to a change in state matter whether that be solid to gas or solid to liquid or gas to liquid or liquid to gas. Any of these and the rest are changes of one state to another. Hence, it allows us to see when the particle changes different states.

rhettfarmer-3H

I think you can tell by the pressure difference outside the thing. Another way to tell is that isometric which is a change in temperature is 0, therefore, it says the reaction is reversible. Also, that reversible tend to do more work which also gives you hint .

rhettfarmer-3H

Yes, exothermic reactions will always have a negative H. However, if the Recants acclimate then it can become spontaneous which I don't know if that shows more about h or the G.

rhettfarmer-3H

As for this one look at the moles of gas on the reactant and product side. Since T and R and are constant, we just want the moles. If moles of products are more than reactants because nf -ni is the change.

rhettfarmer-3H

Hey, Quinn Sprague, I think you miss read the question there is very bold information your missing. It says in the problem that cp =4r and we know from the equation that cv=4r-r.
Other than that use mcdeltaT .

rhettfarmer-3H

I know that Cp is always greater than Cv. From that Cv<Cp. So Cp-r=Cv.

rhettfarmer-3H

Can someone explain the significance of reversible and irreversible? what different equations apply to each other and why ?
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rhettfarmer-3H

I think q=-w when the work is being done by the force of the system. Hence, it takes from the energy, therefore, its q+w and when subtracted to the other side it becomes q=-w.

rhettfarmer-3H

Q refers to heat which is not a state property and H is the enthalpy which is a state function.

rhettfarmer-3H

Liquid to gas would be heating it. The phase changes are important because it shows the importance of temperature on k because it shows that how heat can change the state of the reaction and product which cause different H's. so this realtes to equilibrium and such.

rhettfarmer-3H

the recants are being broken down and the products are being formed. So, exothermic is the formation of bonds, and endothermic is the breaking of bonds. So, when looking at bond strength which every side that has stronger bonds is going to result In more energy released or absorbed. Therefore, you c...

rhettfarmer-3H

I think the best way to think of the reason steam causes more damage is by looking at the heat change of water graph. Hence, when we look at it the phase change to L to G is way higher in temp and energy than S to L. Therefore, when steam hits you more energy is used to lower the temp back down the ...

rhettfarmer-3H

yes, I think it is common that positive H is always endothermic and exothermic is always negative. There is no real expectation of these rules because H will always change based on the environment and changes.

rhettfarmer-3H

The Le chatelier's Principle is essentially a migration of pressure on the solution. In this idea, the reaction will do the least to mitigate the harmfulness of the change of environment. Henceforth, it is the principals of how reactions respond to such changes.

rhettfarmer-3H

pka is the -log of the Ka. Therefore, pka gives us the ph of the Ka solution. Therefore, it acts the same way the ph scale works.

rhettfarmer-3H

I think he was saying that if you are deprotonating an HA and if the ph goes up like ph=8 it actually isn't. Because it would make no sense of ph to go up. And therefore we just rationalize that the ph is just neutral. So there is a common-sense factor when doing these problems and he blames the bre...

rhettfarmer-3H

For the strong acids and bases, there are most of them which you have to just purely memorize which can be memorized. Other than that remember the oh (carboxyl groups) are commonly found as acids and groups 1 and 2 oxides are basics. Cations are usually basic and anions are usually acids.

rhettfarmer-3H

The best way to do this is to set an ice table and the initial concentration of HA is .21M and on the right of it is h30+ and A-. They both start with 0. Then you add x to both products and subtract from X from the initial HA. Then write the ka equation =x^2/(.21-x) because ka is less than 10^-4. we...

rhettfarmer-3H

I also had a hard time answering this but now I think the way I understood it was that Na+ does not affect the ph of the solution so it is not mentioned in the whole of the systematic equation. Furthermore, Na dissolves in water and does not interact with water to create rippling change.

rhettfarmer-3H

Yeah, so it's basically the best way to describe it is that for acid if PH is lower than PKA that means that the PKA (the weak acid) the weak acid won't deprotonate because the PH is lower than the acid so neutral. Whereas if the PH> PKA the weak acid will deprotonate meaning that it will give off i...

rhettfarmer-3H

Percent Ionization =What you have made conjugate acid or base /initial base or acid *100. Therefore first you need to find all the ingredients for it. So first set up the equilibrium table and find the values. then plug them in. Hence, I look at it as a post percentage of how much was converted.

rhettfarmer-3H

If we added more to the reactants it is basically the same just the initial amount would change. Set up the equation as normally and have the right initials and the math will be different because we have different numbers and R will be more than P. Hence, we will just get more product because R>P an...

rhettfarmer-3H

It only applies to gasses so don't worry about in another context. Also, it follows the least moles. One of the crazier ideas is if the moles are the same is that there is no direct change rather it just decrease equally.

rhettfarmer-3H

Q is basically K at a time of the reaction that is not an equilibrium. Use the term Q to basically indicate that it is not at equilibrium. Therefore, they are entirely the same, it helps to describe the state of the reaction.

rhettfarmer-3H

in a short way, Kw is the Kc when water breaks down into H30 and OH. Therefore, this is extremely important. It creates the PH and POH scale it also allows us to convert from KA to KB and visa versa when we need it. SO therefore it is very important and it defines the whole base and acids concepts. ...

rhettfarmer-3H

Yes, delta H is a perfect and most direct way to tell if it's endothermic or exothermic. endothermic being negative(absorbed) and exothermic positive(released). Another way to tell is where the heat is. If heat is in the product then it's exothermic and if it's a reactant it's endothermic.

rhettfarmer-3H

Yeah, q will eventually become K. Q is the product and reaction ratio at a certain time of the reaction. However, as we can infer that it takes a while to get to K. so q is the intermediate until we reach it. Therefore, Q is just a notion of a certain time before it hits K.

rhettfarmer-3H

In terms of the formula, k=P/R. So, as you increase R It would increase P. However I would also like to know whether two reactants play a role as we learned limiting reactions in 14a. However, my basic understanding is that as long as R as a whole increase so P will decrease to balance

rhettfarmer-3H

Not every time does a change in pressure mean a change in concentration, only when compressed or expanded. We Can agree that concentration is made by n/V which means it is dependent on moles and volume. If we compress the system we are changing the pressure but we are also changing the volume. This ...

rhettfarmer-3H

I learned this in LS7A but it has to do with the amount of product and reactants and le chalifer principal. So if Q is really big we have a lot of product. We know from le chalifer that the reaction wants to be at equilibrium. So, it must adapt and produce the reverse so it returns to its stable sta...

rhettfarmer-3H

K=1 means that reactants and products literally have the same concentration. Lavalle describes this as rare because think about how rare that has to be that they are perfectly the same, and in terms of the reaction, it is hard for the reaction to flow and create a product or vice versa. Also, means ...

rhettfarmer-3H

The Le chandelier principles basically are trying to readjust back to normal. So if you remove the product, then there is an imbalance and it is key to remember there are still reactants. So, we still need to obtain that K ratio value, and the K value is constant. So, the product formation is favore...

rhettfarmer-3H

Well to begin it has to be said that Kc refers to concentration and Kp is referring to pressure. With that, the first note is that Kp sorely can be used for gases. Whereas, Kc can be used universally. so one asks when do you pick which one? Well, remember it is based on the equation states and what ...

rhettfarmer-3H

I think the main way to focus on the equation that makes the salt. So, for example, NaCl is made of NaOH and HCl strong base and acid so they cancel to be neutral. As for the others if we have a strong acid and weak basic is a acid salt and another is a if a strong base and weak acid therefore, its ...

rhettfarmer-3H

I would focus on the stiochemistry point on the graph and why it matters.This point will tell you what type of acids and base you have and the ph of the overall solution.

rhettfarmer-3H

The huge difference between the two is that hemoglobin is 4 myoglobin complexes which 1 myoglobin is one heme structure. Therefore, hemoglobin can carry 4 oxygen because it has 4 heme structures. Whereas, the myoglobin can only carry one O2. Hence, there is 4 histones and 4 o2 on the hemoglobin. The...

rhettfarmer-3H

I think it is key to note that the NH3 on the cisplatin has already filled its self and doesn't have any lone pairs or space for it binds. Therefore, when we look at the Cl- we have 3 pairs of lone pairs on both and its is also has a dipole movement. Hence, it is more likely that we bind at this sit...

rhettfarmer-3H

How I like to think of neutral charts is to think of the things that made the salt so I work backward. SO for example, NaCl is a neutral salt. How would I know if it's neutral I'm gonna make the acid and the base that makes the salt. Hence, we have a strong base of NaOH and a strong HCL. How do I kn...

rhettfarmer-3H

So Titrations are not usually known before, Like we said we use it to see if it balanced, acid or basic. We are focusing on that stoichiometric point, Which means that the two moles of bases and acids are the same. This will let you know if the sample is acid or basic. Furthermore, it can help in th...

rhettfarmer-3H

So, because the oxo-acids have the H attached to the O and not the halogen this causes a great difference in the stability of the molecule. Therefore, we look at the induction - which is the spread of the charge across the anion: which I really mean is the stability of the anion. It is key to note t...

rhettfarmer-3H

CO is monodentate for many reasons to begin with the structure of CO-. Its one lone pair on C with a triple bond and a lone pair on O. Therefore, the O is negatively charged. Hence, since the structure is only 2 molecules long there are not enough spacers. Usually, you need 2 spacers because it woul...

rhettfarmer-3H

The suffix -ate is used in the case that the whole complex of the coordination compound is actually negative. Therefore, its anions and hence it will commonly be found at the end of a compound. I say this because the anions are followed after cations. For example Na Cl- therefore, cations than anion...

rhettfarmer-3H

You are right in a way. Octahedral is the most common coordination number 6 geometry for most. The hexagonal planner is not common in most structures because the repulsion is not great for it is a regular structure so that's why it gets ruled out. The others are not coordination number 6 they are 5....

rhettfarmer-3H

yeah for sure, this question is a little tricky in the sense that you have to know you ligands. SO we have 2 different ligands en and we have two of them and we have two CO. For this problem it seems like oh there is simply a coordination number of 4 but wait. En is bidentate so we have to add 2 for...

rhettfarmer-3H

the coordination number is just the number of bonds that the TM has made. So, for your example when we check the we see that next to the TM we have a ligand with F4 so the 4 monodentate ligands make 4 bonds so the coordination bond is 4. It is key to check that they are not bidentate or polydentate ...

rhettfarmer-3H

En is a ligand which is the short name for ethylenediamine. The key thing to remember here is that ethylenediamine or en is bidentate. Therefore, when counting for coordination number count this as 2 ligands. Confusing but check out the ligand page Lavalle has it helps a lot.

rhettfarmer-3H

Hemoglobin is used in our blood to transfer oxygen and myoglobin is used in our muscles to transfer o2 too. Hence, both are made of the same structure the heme structure with the same protein and 02 Hence, one is just more complex than the other. For example, the hemoglobin is 4 myoglobin which comm...

rhettfarmer-3H

I think the best way of thinking of this in this class. Is that the heme structure, the histine, and the oxygen make the complete complex of myoglobin. Therefore, it answer your question the protein is attached to the bottom of the Fe so it is part of the heme structure in a way. So the 4 NH3 and o2...

rhettfarmer-3H

comes from the fact there are four bonds that can be made from carbon and therefore, there are 4 regions of electron density. This correlates to sp^3 because it's a combination of 1s and 3p's which makes 4 regions. Likewise, with the other 2 regions is sp and 3 regions would be sp^2. 4 regions goes ...

rhettfarmer-3H

Sigma is a correlation of one bond. It is also one bond in the double bond and so in always one sigma bond. Furthermore, pi bonds are the spots onwards on the double bonds and triple. the big difference is that sigma bonds can rotate by the feature of them being tied in less rigid way in comparison ...

rhettfarmer-3H

Hi! This is another part of Sapling Q.11. I understood how your answers applied to the other diagram, but this diagram threw me for a loop. If any of you have advice on how to better understand and complete these questions it would be much appreciated! Thank you! for this one there are multiple str...

rhettfarmer-3H

the most prevalent hybridization is the regions of electron density which stems from lone pairs and bonding regions. If you ask for hints it gives you a really good chart that explains which hybridization matches with which orbitals. 2 region---> sp 3 region---> sp^2 4 region---> sp^3 5 region---> s...

rhettfarmer-3H

yes, it is delocalized. I think of it as for the sake of the structure having multiple figures for the double bonds. The overall bond lengths for one of the bonds in each of the carbons have to be different from the rest of the hybridized other bonds. Therefore, these represent that pi bond in the d...

rhettfarmer-3H

Every covalent bonding region has a sigma bond. This is to also add that one bond in a double bond is a single bond and the other is sigma. Also, there is no difference rather it describes the character of a single bond. It also can rotate, unlike PI bonds.

rhettfarmer-3H

yeah, iconic radius is dependent on a charge. Think of it as when something is a more negative or more negative charge that means there are more electrons in it. Also, in iconic compounds. Follow the trend lavelle gave us that bigger molecules are going to have more radius so as we move down on the ...

rhettfarmer-3H

yes, this is very common and it has to do with dipole movements. An easy one is C02. co bonds are polar bonds. but for co2 its two double bonds on either side of the c. Hence, they both move their dipole towards the c so the movement is actually counteracting each other and therefore, it is even and...

rhettfarmer-3H

Phosphate can have an extended octet because it is the first row of p block before the introduction of the 3d block. Therefore, it is open to be used. Hence, that is why phosphorus and elements in the p block below the 3 p are able to open and sue the 3d block. Also, it is low energy so it's not too...

rhettfarmer-3H

Yeah, they can a common one is S. S is commonly seen with over 10 valence electrons. SF6 has 12 electrons on the one sulfur molecule and it is seen to be found and a molecule so yes.

rhettfarmer-3H

to figure out which is more iconic it depends on what the molecule is made out of. For example, more iconic is going to have smaller anions like fluorine because they are less covalent and more negative charges. Whereas, for cations, they are gonna be bigger such as K instead of NA because the small...

rhettfarmer-3H

For melting and boiling points lavelle talked about polarizability being the main factor. So, the common element that add to polarizability is the size of the molecule. So when you compare two elements look at the size. when I mean size atomic radius is a key trend or number of electrons.

rhettfarmer-3H

I think a big part of the question has to do with polarity. Therefore, it has to be shown that if you its polar it can make the dipole connections. SO there no polar molecules that are likely to have only London. And London happens in all molecules so the answers are BCL Br2 and c2h6

rhettfarmer-3H

Hey, I think the point of focus is on the lone pairs. You were right about the other pairs being possible but the two you may the miss were the lone pair on nitrogen and the 2 lone pairs on oxygen. I missed this part too. It also depends on lone pairs. Since, o has 2 there are two bonds that can for...

rhettfarmer-3H

I also found this idea questionable. I had a very hard time. However, the best way to think of it is we are trying to minimize the formal charge to create a neutral atom. So, with resonance, the best way to get the best answer is by having the lowest formal charge more neutral more stable more likel...

rhettfarmer-3H

I think for this question it is very important to remember that there are certain rules that must be remembered. One of which is the energy levels the D block has more energy than the s block therefore, when we fill the d block rather than filling the s block the element becomes more stable. Similar...

rhettfarmer-3H

Since fluorine has 7 electrons it has a uncatering ability to get to that filled valence electrons therefore, it battles for that extra one electron and is willing to go the mile to get it, it makes sense that it wants that last electron. Also it for that sake it has the highest in its column becaus...

rhettfarmer-3H

Thank you for posting this after reading Chem mods discussion this is what I retain because I am also having a hard time understanding. That formal charge is due to the stability of the element/molecule. hence, the FC is to be lowered for the sake of stability. And formal charge gives certain charac...

rhettfarmer-3H

Atomic radius decreases as we move to the right of the periodic table because more electrons we have a closer outer shell because of the attraction of the inner nuclear charge is increasing at the same right while the shielding of electrons is remaining the same so there is a short radius by a stron...

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