CHEMISTRY COMMUNITY

Rohit Srinivas 2D

For this one you will need to figure out which step is the rate determining one. That is the slowest step. The experimental rate law should clue you in on this. It implies that it has to be a step where H2 is reacting with another molecule. However step 2 has this but has n2o2 instead. This is where...

Rohit Srinivas 2D

Temperature can actually change A, This can be seen by re organizing the Arrhenius equation. This website has an excellent explanation of how https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/06...

Rohit Srinivas 2D

I solved the cell using two half reactions that the answer key did not use. I ended up with the correct cell potential and the correct overall equation, yet the answer key uses different half reactions and hence they reach a different cell diagram. Is my cell diagram correct as an alternative proces...

Rohit Srinivas 2D

I think it really helps to follow the procedure to a T. The procedure is outlined in the textbook on page 538.

Rohit Srinivas 2D

Oxidation number rules can help with this. You know that halogens will be -1, and that alkali will be +1. This means those substances are highly unlikely to change.

Rohit Srinivas 2D

Adding on to the reply above, the substance needs to be connected to the structure you are trying to protect. This connection needs to be conductive, and in essence the cathode is sacrificed to save the structure.

Rohit Srinivas 2D

Adding on to the reply above, you also need to know how to calculate cell potential from a list of standard reduction potentials.

Rohit Srinivas 2D

Unfortunately, that means your calculator cannot compute such a large number. I do not think there is a way around that. Apparently calculators cannot handle numbers above 9.99999999 x 10^99.

Rohit Srinivas 2D

E naught is not equal to zero at equilibrium I believe. It is only when K is equal to 1 that the whole equation is equal to 0. Not all equilibria have a K value of 0 only very few do. Hope that helps in addition to the reply above.

Rohit Srinivas 2D

hello everyone,
here is an interesting application of the thermoelectric effect that I found (due to all the news about rovers)
https://rps.nasa.gov/power-and-thermal- ... r-systems/
I was wondering if anyone knows how to calculate the cell potential for one of these units?

Rohit Srinivas 2D

Why is E naught not dependent on moles? Wouldn't the coefficients dictating how much electrons can move lead to some sort of a limiting reactant in the one that uses up more electrons per reduction?

Rohit Srinivas 2D

E is negatively correlated with delta G. When E is negative delta D is positive which means that the situation is not spontaneous. When E is positive, delta G becomes negative making it spontaneous. This is seen from looking at deltaG= -nFEnaught

Rohit Srinivas 2D

It is also known as a selective ion electrode, which basically tracks concentration of a specific ion. A common example of this is the glass electrode, which is used to track pH changes.

Rohit Srinivas 2D

We can use log or ln. We just have to use different coefficients. I believe that is what your question is asking?

Rohit Srinivas 2D

I believe that we are not going in exact order, so probably just read the textbook sections when we cover them in class? I am personally on 6M.

Rohit Srinivas 2D

Salt bridges and porous membranes serve the same purpose. However sometimes there are errors in either set up as seen in this paper https://scholar.harvard.edu/maralmousav ... rous-glass

Rohit Srinivas 2D

As the replies above seem to indicate. The thermodynamic properties are extensive properties, which means they depend on the amount of substance. Reduction potential is not extensive.

Rohit Srinivas 2D

David Y wrote:Cp is 5/2R while Cv is 3/2R

yeah I was wondering why there was a difference of R, how does that arise from having a fixed/changing volume?

Rohit Srinivas 2D

Another important equation is Delta S =nR ln(v2/v1)

Rohit Srinivas 2D

You use avogadros number when you have a mole of a certain molecule. This would make degeneracy raised to the power of avogadros number to account for each of the positions on each molecule.

Rohit Srinivas 2D

The first one has 4 *3 positions. This is because cis means the red atoms must stay next to the other. Hence in the middle section there are 4 slots for it, then 4 that would work from the top to the middle and 4 that work from the bottom to the middle. As for the second one there's only 6 atoms so ...

Rohit Srinivas 2D

Could someone explain how the mathematical relationship between Cp and Cv was derived? I was looking at it in the textbook and got confused.

Rohit Srinivas 2D

Why do we use Cv in the delta S=nC ln T2/T1 formula and not the Cp value?

Rohit Srinivas 2D

This is because a reversible reaction assumes a perfect system, with no energy lost other than work. This is the theoretical limit of a system.

Rohit Srinivas 2D

work of expansion, because the external gas around the piston is expanding, even if the gas inside the piston is compressing.

Rohit Srinivas 2D

The first equation has atm that you use the appropriate R for. The second R is only for Joules by Kelvin. The units dictate what R value you use.

Rohit Srinivas 2D

The system is doing work, which means that some amount of internal energy is being lost. But temperature isn't falling, this implies that some other energy is entering the system to keep it at thermal equilibrium. Hence, it is a heat sink.

Rohit Srinivas 2D

I think we multiply by RT or R to figure out the H and delta S, or they might be asking us to analyze a graph.

Rohit Srinivas 2D

The above response is excellent. I just have an example to add. Check out Example 4d.4 to see what an alternative fuel problem looks like.

Rohit Srinivas 2D

You would have to find a scenario in which enthalpy is either very large and where entropy is very small (which the above replies have covered). This is because a large enthalpy value is not likely to be skewed too much by a change in entropy. So a large negative enthalpy would correlate to a large ...

Rohit Srinivas 2D

Disorder has become synonymous with the "wrong/bad outcome". Entropy does not follow the value that media and society has assigned with that word. Instead entropy can just be thought of as randomness in a system.

Rohit Srinivas 2D

How are absolute enthalpy and absolute internal pressure different than enthalpy and internal pressure? Absolute enthalpy and internal pressure are the total amounts of those quantities. The values we use in our problems are a change in enthalpy and pressure. Those are observed by experiments where...

Rohit Srinivas 2D

Can absolute enthalpy and absolute internal energy be determined? I believe that absolute enthalpy of a system cannot be measured, however it is possible to estimate this using the change in enthalpy observed. As then the energy would be used to calculate how much energy would need to be taken out ...

Rohit Srinivas 2D

One thing to add to the previous answer would be to consider friction. The friction between the piston and the wall could be another potential avenue for the energy to leave the system and enter the surroundings.

Rohit Srinivas 2D

Enthalpy change is usually calculated as kilojoules per mole, as when we are looking at an equation. However, if we want to know the q or the change in a system we would need to multiply it by moles to get the kilojoules that occurred in a specific reaction. So in summary, use kj/mol if we are doing...

Rohit Srinivas 2D

We do not need to know how to solve calorimetry problems for the midterm, but I believe we will learn them soon.

Rohit Srinivas 2D

We would need to be given the change in enthalpy or we would need to conduct the experiment in a calorimeter to track the change in the temperature. We do know that q= MCAT which can be used to calculate the change in heat energy.

Rohit Srinivas 2D

We will need to know the heat of fusion and vaporization, which I believe will be on the constants sheet. Additionally, I believe it is not like the photon, as it has a lot of different factors. Sometimes water can be chilled below the temperature as it will need a freezing nucleus to form ice. At t...

Rohit Srinivas 2D

Is the state of a system a state property? During a phase change the system at a particular enthalpy can exist in either phase, which would mean that state of matter is not a state property as it depends on the pathway it took to get to the phase change temperature. Additionally isn't the enthalpy a...

Rohit Srinivas 2D

The reply above does an excellent job explaining it but I believe one more detail could help make it clearer. The reason why the neutral ph changed is because Kw has changed as temperature changes move the equilibrium constant.

Rohit Srinivas 2D

I believe if you understand how that equation is derived (from the ideal gas law) then you should be fine. We were supposed to learn the ideal gas law from the outlines, so I would assume that these kinds of questions are fair game for the midterms.

Rohit Srinivas 2D

Would we ever need to consider the volume a solid reactant or liquid reactant takes up in a reaction chamber? For instance if we added a large excess of a low density solid to a reaction vessel, wouldn't that change the volume and thereby pressure of the system for which we would need to apply le Ch...

Rohit Srinivas 2D

This is peculiar because our bodies are slightly alkaline. Luckily we have a nice system of buffers, which we will probably learn the Henderson hasselbach equation for, which keep our bodies at a ph of 7.4.

Rohit Srinivas 2D

Yes, this is because ideal gasses assume that they do not interact with each other as their collisions are considered negligible. Hence, their partial pressures are calculated which is equal to their "concentration". Additionally, the units for Kp do not matter as they will cancel out if t...

Rohit Srinivas 2D

I believe for 6B.9 you do still use the kw of 10^-14.

Rohit Srinivas 2D

They would change but not by much, which is why we disregard them. The change is so small it doesn't affect the calculation we make.

Rohit Srinivas 2D

The pressure could increase from an inert gas, that would not affect the K value.

Rohit Srinivas 2D

Sometimes you cannot solve for the concentrations at the start so you have to do it at the end. This was what I found from problems on sapling. This is because we are only given initial concentrations not equilibrium ones, hence we could only solve for it at the end.

Rohit Srinivas 2D

RICE is the same thing as an ICE table, my high school used both names for it. I think just knowing how to solve the problems is important.

Rohit Srinivas 2D

An inert gas does not change the volume. K is always calculated in terms of quanitity per volume, hence increasing pressure with an inert gas does nothing to change it.

Rohit Srinivas 2D

What are your thoughts on this?
https://www.theguardian.com/science/2016/nov/28/theory-challenging-einsteins-view-on-speed-of-light-could-soon-be-tested

Rohit Srinivas 2D

We use the henderson hasselbalch equation: pH=pKa +log ([A-]/[HA])

Rohit Srinivas 2D

We add in molarity (concentration). We haven't done kinetics yet but whenever we see the [] that indicates concentration.

Rohit Srinivas 2D

The response above talks about group 1 and two. So I will talk about some transitional metals. These also can form hydroxides and oxides as seen in the textbook. This is because they can become a +3 or +4 ion and can bind to oxygen which is -2 or hydroxide which is -1

Rohit Srinivas 2D

I believe that this is referencing the fact that strong acid and bases in water would dissolve, hence would not be written in molecular form but as ions. Weak acids and bases are only dissolving at a low rate which is why their molecular formula is utilized.

Rohit Srinivas 2D

I believe the reason concentration doesn't matter is because when the concentration is very high there won't be enough water to be protonated and participate in the reaction, at a dilute concentration there will be enough water to do so. Also both bases and acids are corrosive, not just acids.

Rohit Srinivas 2D

I believe we should be memorizing the chart on page 724 at the very least. That has a lot of common ligand names.

Rohit Srinivas 2D

Well as we know the strong acids, and none of those seem to be amphoteric, it appears that most amphoteric molecules that can act as acids would be weak ones. Same would apply to bases.

Rohit Srinivas 2D

As the above reply indicates, having more carbon dioxide leads to more carbonic acid which lowers the ph of solutions. Respiratory acidosis is when the lungs cannot expel enough CO2 that it leads to the blood and bodily fluids becoming acidic.

Rohit Srinivas 2D

A pH of 7 is neutral. 0-7 are considered acidic. 7-14 are considered basic. This is because it is the negative logarithm of the concentration of H+

Rohit Srinivas 2D

Please tell me everything you know for describing the structure of a molecule, cation, or anion in terms of hybrid orbitals and sigma and pi bonds! - To describe the structure of a molecule, cation, or anion, you will need to know the amount of lone pairs on the central atom and the amount of bonds...

Rohit Srinivas 2D

Lone pair repel each-other more greatly than two atoms, hence they increase the angle between themselves, which in turn pushes the two atoms closer together decreasing that angle. This results in the bent shape.

Rohit Srinivas 2D

This is because a pi bond is weaker than a sigma bond. The first bond that always forms is a sigma bond. Additionally pi bonds happen at 90 degrees to each other (think of the p shell orbitals) and that is physically impossible to happen and be bonded to the same atom.

Rohit Srinivas 2D

I believe this is figured out by looking at pi bonding. whenever you have an even number there are enough pi bonds to lock the carbon into a single plane. This is because the pi bond does not allow for rotation. Every pi bond is formed at a 90 degree Angle to the other, so even numbers would go back...

Rohit Srinivas 2D

Hybridization is the mixing of shells to form half full orbitals that can bond at an energy level between the original energy levels of the constituent shells. This is because of the Pauli exclusion principal that states you can only have 2 electrons in a shell. In response to your second question, ...

Rohit Srinivas 2D

For the scope of this course (as far as I can tell from the textbook problems) we just draw out all the possible resonance structures then label the bonds for each structure. As for the real world, I think that the Pi bonds are the reason why resonance can exist. They are easily broken (relative to ...

Rohit Srinivas 2D

You can do this by looking at the ionic character of the bonds. So a c-F bond is more polar than a C-o bond. So the vectors for that molecule will have a stronger partial negative charge. Furthermore, if the vectors in a molecule cancel out, then it would be less polar than one where the vectors do ...

Rohit Srinivas 2D

I believe that is correct, you put it on the nitrogen. When you have the option that all else has the lowest formal charge and the only difference is where a negative formal charge goes, then you put that on the more electronegative atom. I believe this is similar to a sapling problem from last week...

Rohit Srinivas 2D

Their electronegativity difference is about 0.4 which means it is non polar. Anything above 2.0 is considered ionic bonding. Since 0.4 is so small it is considered non polar as the ionic character of this bond is very low.

Rohit Srinivas 2D

It is not possible for the S orbital to overlap in that manner. If you look at it geometrically, it is the same at all aspects, hence an overlap "sideways" is the same as the overlap from a sigma bond. This means that they can only create one bond.

Rohit Srinivas 2D

The bigger the atom is the more polarizing power it has, because it has more protons pulling on the electrons. Hence, Be would be the smallest and lowest, then Rb, then Sr. This means polarizing power goes up across a period and increases as u go down a group.

Rohit Srinivas 2D

I believe the question is asking which one of the molecules has hydrogen bonding in water. (these are the atoms with H bound to F, O, or N) HF is the only molecule that meets this criteria

Rohit Srinivas 2D

The above reply explains how to do it, so I will help with when to do it. You can do it when the atom is in the 3p shell onwards as it can hybridize with the 3d shell and accept electrons there as it is technically also at the third energy level. You would do this when you need to put in more electr...

Rohit Srinivas 2D

Can a hydrogen bond form with F 2 or not? Additionally if a Flourine atom was in a polar molecule would it be able to participate in hydrogen from another atom that has an oh or nh or fh group, despite the fluorine in the polar molecule not being bound to hydrogen (lets suppose its bound to carbon o...

Rohit Srinivas 2D

I noticed during some of the homework on sapling the hydrogen bonds could be created twice on the same oxygen atom because it has two lone pairs. Are hydrogen bonds like coordinate covalent bonds, or does it have to do with the fact that each lone pair simply acts as a negative charge for positive d...

Rohit Srinivas 2D

This is outside of the scope of this class, but it was cool so I searched it up. There is a person named Linus Pauling who made a scale to measure electronegativity from 0.79-3.98 (1-4). He assigned Flourine to be 4, and then calculated the rest through experimental observation. This is known as the...

Rohit Srinivas 2D

I believe most replies above have the definition of London dispersion forces. Here is the order of bond strengths from strongest to weakest
1. Ionic bonds
2. Covalent bonds
3. Dipole Dipole bonds
4. Van der Waals (London dispersion)

Rohit Srinivas 2D

Delocalized electrons are free flowing between the nuclei. We just simplify it for our Lewis diagrams. If you have heard of metallic bonds, those are an excellent example of delocalized electrons, and the reason we have electricity!

Rohit Srinivas 2D

Covalent happen between two nonmetals. Additionally, the atoms in ionic bonds with covalent characteristics have a higher electronegativity difference than atoms in polar covalent bonds.

Rohit Srinivas 2D

I believe he specifically highlighted them because they are insoluble. Hence this physical phenomenon we can observe with our eyes shows the covalent characteristics. This is unexpected for an ionic compound.

Rohit Srinivas 2D

The answer is c. This is because it deviates from the normal rules for filling orbitals, you can notice how one orbital at a lower energy level is half filled and instead the electron is in a higher energy level. The rest of the options follow hunds aufau and Pauli

Rohit Srinivas 2D

You can calculate it by understanding the ml quantum number. this tells you how many orbitals are available, each orbital has 2 so it would be ml options *2. If we look at the shells we know so far: s has only 0 as ml so 1*2=2 p has -1,0,1 as ml so 3*2=6 d has -2,-1,0,1,2 so 5*2=10 f has -3,-2,-1,0,...

Rohit Srinivas 2D

I looked at the periodic table. How many electrons are possible at the second energy level are all of the atoms in the second period. this would be Li-Ne. Or aka 2s^22p^6. This is 2+6=8

Rohit Srinivas 2D

This all depends on energy level, obviously the 5s electrons will have more energy (be further away from the nucleus) than a 4s electron. However when comparing orbitals at the same principle quantum number, the order will be s, p , d ,f. This has to do with the physical geometry determined by shrod...

Rohit Srinivas 2D

The way I memorized them is by looking at the periodic table. I have memorized the shape of the "staircase" and can reproduce it from memory. This method is probably less efficient than a mnemonic but goes to show there are alternative ways of memorizing them

Rohit Srinivas 2D

Tl;dr If the question asks about the energy of the electron it will be negative as it loses energy. If the question is asking how much energy is released or is in the photon released that is positive.

Rohit Srinivas 2D

On one of the textbook problems it assumed we knew how to convert from miles to meters. Is that something we need to memorize?

Rohit Srinivas 2D

I was looking at the figure 1b.1 in the textbook which depicts black body radiation and energy per wavelength of light. How is it possible that objects at higher temperatures emit a higher amount of energy for a certain wavelength than a cooler object emitting photons at the same wavelength. Is this...

Rohit Srinivas 2D

Here is how I approached this problem. First I converted from wavelength to frequency using hv=c. v= 1.577x10^15 Then I converted to energy per photon using hv=E. E= 1.045x10^-18 J then I converted to ev. this gave me 6.53 eV. I subtracted kinetic energy to find the work for the metal. This is 2.53 ...

Rohit Srinivas 2D

I believe what you are referring to is the Rydberg equation. V=R (1/n(final) - 1/n(initial)) will always give you a positive number. Think about it in terms of fractions, final will always be smaller than the initial. The larger a denominator, the smaller the fraction. Hence, 1/small - 1/large will ...

Rohit Srinivas 2D

Balmer ends at n=2, Lymann at n=1, the Paschen series ends at n=3 and the Brackett series ends at n=4. These are the ones from the textbook.

Rohit Srinivas 2D

We don't need to know this but something interesting to know is that T is temperature, C is the same C we have been using, and lambda is wavelength. This is much more advanced that we would be learning for 14a.

Rohit Srinivas 2D

Yes, that is exactly correct. Conceptually, the equations we are solving are PER photon. So at the end we would just need to figure out how many of the energy per photon we would need to get the 32.74J. First you would want to find the frequency so do c=lambda v this would give you 3.08 x 10^11 hz T...

Rohit Srinivas 2D

I believe n1 is the final. This is because n1 being the final would mean its the lower energy state, and the n2 term would have the higher energy state. - Small- (-big) would create a positive energy, which tells you the amount of energy released. This is exactly the same as Dr.Lavelle explained. I ...

Rohit Srinivas 2D

How would we calculate the momentum of a photon? Is it a similar calculation to that of an electron except we use the mass of a photon instead? What would be the units? thanks.

Rohit Srinivas 2D

As most replies above have indicated, the speed of light is a constant and will be provided on exams. As for why that is? this is where it gets interesting. The speed of light is not given to us in a relative sense, as a car passing by us may appear to be 40 mph but only 0 mph for those inside of it...

Rohit Srinivas 2D

As the reply above indicates, 3000 would not be considered an exact number unless you have a decimal. Hence, 3020 would have 3 significant figures, as it could be written as 3.02 x 10^4. The general rule is digits after the decimal place are considered significant, and trailing zeroes without a deci...

Rohit Srinivas 2D

This should be either given or located inside the textbook. All these activities are simply examples on the electromagnetic spectrum.

Rohit Srinivas 2D

I believe it goes from n-6 to n=1, so that would be 5 spectral lines. One for each transition it makes, as the energy we observe is released when the electron moves down energy levels.

Rohit Srinivas 2D

This is an interesting question! Here are my thoughts : 1. As numerous others have mentioned in this thread, getting to a 100% actual yield will be very hard. This assumes every atom in the reactants reacts in the appropriate reaction that you are investigating. 2. You might get a false 100% yield a...

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