CHEMISTRY COMMUNITY

AtreyiMitra2L

When there is no solid metal in a reaction, add Pt(s).

AtreyiMitra2L

Adiabatic is when there is no transfer of heat. Iso thermal is when the change in internal energy is 0.

AtreyiMitra2L

Aldehyde is on the end and ketone is in the middle.

AtreyiMitra2L

For the salt bridge, we use the double line and for the porous disk we use a single line.

AtreyiMitra2L

Everything but 15.9.

AtreyiMitra2L

In general, have the anode go first and the cathode go next. The inert metals should go on the ends. For those in the middle, don't worry too much about the order as Dr. Lavelle has said that it is not that important.

AtreyiMitra2L

Generally, the heavier the substance, the higher the molar entropy. This occurs because the energy levels are closer together, which means there are more possible orientations the substance can acquire. Therefore, it will have a higher molar entropy.

AtreyiMitra2L

I basically look at what reactant is being oxidized and name that the reducing agent. Then I look atwhat reactant is being reduced and name that the oxidizing agent.

AtreyiMitra2L

furthermore, it is important to note that for isothermal, reversible processes, the change in internal energy is 0 when they are ideal gases. Because they are ideal gases, there is no exhange in energy during the collisions.

AtreyiMitra2L

Enzymes, by defintion, are catalysts. Therefore, they work by speeding up the reaction and not by slowing down.

AtreyiMitra2L

Because we know its constant external pressure, we automatically know that its undergoing an irreversible expansion.

AtreyiMitra2L

Yes. An intermediate is just a substance that is produced and used in the reaction in a manner that it isnt shown in the final reaction. Because this occurs, it will be an intermediate even if it is isnt shown in step 3

AtreyiMitra2L

We can determine the intermediates of the reaction by looking at which cancel out throughout the reaction. By doing this, we can also figure out the number.

AtreyiMitra2L

Hi! Im posting here about this because Im desperate LOL. I am currently enrolled next quarter in chem 14BL with casey in lab 1j (Thursday 6:30 pm - 9:20 pm). I am looking to exchange with anyone from casey or the other professor for a different time lab with chem 14BL. Please comment below if intere...

AtreyiMitra2L

Yes this is the general rule

AtreyiMitra2L

We can include them but its . not needed.

AtreyiMitra2L

You need the table to identify 0 order reactions

AtreyiMitra2L

I believe so.

AtreyiMitra2L

It just the standard because we can only calculate the change rather than the actual value

AtreyiMitra2L

Have the anode then cathode. The inert metals should be on the outside. besides that dont stress

AtreyiMitra2L

In aqueous solutions, water can be oxidized or reduced as well.

AtreyiMitra2L

For basic solutions, to balance O's we use water. Then to balance H's, we use OH's and H2Os on both side.

AtreyiMitra2L

On the right, you see two hyrdroxide ions. This means that there is -2 on that side. To balance the equation, you need negative 2 on the other side. This is why we have 2 electrons.

AtreyiMitra2L

We will have to know how to do both. The list of constants and equations is on his website.

AtreyiMitra2L

but even then, when you look them up there seem to be several possible values. How do you know which one to use.

AtreyiMitra2L

In question 14.21a, we have mercury. Mercury, while a metal, is liquid at room temp. In the answer key, we dont have a inert metal there. Do we use inert metallic components for all metals?

AtreyiMitra2L

Delta s total will only be 0.

AtreyiMitra2L

I just think the oxidation agent is the substance being reduced on the reactants side and the reducing agent is the substance being oxidized on the reactants side.

AtreyiMitra2L

You have to split it up into half reactions. From there, try to see whether oxidation or reduction is occuring. If oxidation is occuring, we have the anode. If reduction is occuring, we have the cathode.

AtreyiMitra2L

How did we calculate heat and work in number 4 on the midterm?

AtreyiMitra2L

If there is net increase of gaseous moles --> increase in entropy. If there is a net decrease --> decrease in entropy. If appearence of aqueous, net increase. if we see an increase in molecular complexity with the products --> net increase.

AtreyiMitra2L

We often use 0.0821 when calulating with PV = nrT. For the others, it is possible I suppose.

AtreyiMitra2L

They are relatively interchangeable when there is only one mole. But if two mole of the product is formed, if you have it as kj per mole, you would need to multiply your value by 2.

AtreyiMitra2L

This can only occur in isothermal, reversible reactions. It would not be considered a violation of the second law of thermodynamics (law that says the entropy is always increasing) because these never actually occur in nature.

AtreyiMitra2L

Do all reversible reactions occur isothermally?

AtreyiMitra2L

we do not need to know this. dont worry about it.

AtreyiMitra2L

We use the former with constant pressure and the latter when its not constant pressure. The first is irreversible expansion and the latter is irreversible, isothermal expansion.

AtreyiMitra2L

W is degeneracy. it doesnt have units. You calculate it by doing the number of orientations ^ (num of molecules)

AtreyiMitra2L

If it is in the same direction, you know there is only one possible orienation. Therefore, regardless of how many molecules, the entropy using the boltzman equation will still be found to be 0. This occurs because the W in this equation is found by (# of possible orientations) ^ (# of molecules). In...

AtreyiMitra2L

The lecture slides are in the course reader than he no longer sells. Getting this would be your best bet.

AtreyiMitra2L

When delta g is 0, it just means that it is at equilibrium. It is not enough to assume the other two quantities to alse be at 0.

AtreyiMitra2L

At the anode, oxidation is occuring. At the cathode, reduction is occuring. If you know these concepts and look at the diagram enough, you will be able to distinguish the two

AtreyiMitra2L

I think its preferable to use KJ but you could use either one

AtreyiMitra2L

When the question says the reaction is occuring at constant pressure, can we use C instead of C (p,m) in our equation (change in S = nC(p,m)ln(T2/T1)? Thanks!

AtreyiMitra2L

Why do substances with larger masses have larger molar entropies?

AtreyiMitra2L

The molar entropy is basically the entropy per mole of a substance. If you want to find the molar entropy, you need to know how many possible positions the molecules has as in number 25. Then, you have an exponent with Avagadoros number. You plug all this in the boltzman equation and you should be g...

AtreyiMitra2L

On the test, you will be given the table of values that will help you solve for cP

AtreyiMitra2L

The book says that water in the liquid form will be formed only with organic compounds. When it is not an organic compund, then gas will be formed.

AtreyiMitra2L

Bond enthalpy is the amount of energy to break one mole of the substance. On the other hand, enthalpy of formation is the amount of energy required to make the substance by substances in their standard states (the state at which they will be found at 1 atm). H2 at one atm is in gas. Therefore, becau...

AtreyiMitra2L

To look for the net production of gas, you simply subtract the moles of the gas on the product side by the moles of gas from the reactants side. The one mole of OF2 is just refering for the final answer to be in per mole.

AtreyiMitra2L

I believe it is deltaU = deltaH - delta(n)RT. This goes off of the concept that delta u is equal to q + w. (AKA the second one). Work is the second one. This occurs because of the idea of what the work is being done on or by what. In expansion, the work is being done ON the surroundings BY the syste...

AtreyiMitra2L

Why is it that in isothermal, reversible expansion heat can be done if heat is defined as the transfer of energy due to a temperature difference? I understand that the change in internal energy must be 0 for ideal gases so heat must be involved but I dont understand it past this. Thanks!

AtreyiMitra2L

Well, usually for irreverisible, work = -Pex * (delta V). I am not sure where your equation came from. I believe in this class for reversible, isothermal expansion, all you have to worry about is the one with natural log.

AtreyiMitra2L

In hess's law, we are calculating the overall enthalpy change. Because enthalpy is a state function, we know it isnt pathway dependent. Therefore, it doesn't matter which way it got there. If you have fractions, as long as you changed enthalpy along with it its fine :)

AtreyiMitra2L

R is related to the gas constant or 8.3145 J/ (K x mol). It will just be a constant.

AtreyiMitra2L

For bond enthalpies, I believe you got the order wrong. It should be reactants - products. This idea is built on the idea that to break bonds in the reactants. To build the bonds in the products, there is a release of energy. Therefore, there is reatants - products.

AtreyiMitra2L

Hello! When a liquid combusts, will liquid water be produced and not gaseous water? Does this always happen? I am asking this specifically in reference to num 93 in which the combustion of c6h6 seems to produces liquid water. Thanks!

AtreyiMitra2L

In a closed system, there is an exchange of energy but not matter. The example in the book given is a refrigerator coil. This is a closed system because the coil cools the refrigerator through the transfer of energy. There is no exchange of the fluid, however, In contrast, in an isolated system, the...

AtreyiMitra2L

This concept goes back to 14A. In reasonance, delocalization occurs. Because the electrons are spread over a wider area, the resulting substance is more stable because there are fewer electron electron repulsions.

AtreyiMitra2L

I think because we are supposed to be looking at it in terms of the absolute value. In chemistry, we always take the system's perspective. In physics, however, we usually take the surrounding's perspective. Because of this discepancy, we look at it in terms of absolute value. Please don't hesitate t...

AtreyiMitra2L

Yes you can! These properties are all possible because enthalpy is a state function. In other words, it doesnt matter how it got there. The end result will still be the same.

AtreyiMitra2L

If you think about it, the more atoms there are, the more heat is required to raise the substance by 1 degree celsius. Therefore, because NO2 has 3 atoms rather than only 2, it takes more heat because there are more atoms that can absorb the heat.

AtreyiMitra2L

Why is it that heat at constant pressure (enthalpy) is considered a state function but not heat? Thanks!

AtreyiMitra2L

For it to be the most stable form, it must be the one that is most pure. Some of them you just have to memorize. But there are some that you can figure out. All noble gases are most stable in the element phase as a gas. All diatomic gases (such as N2, O2, Cl2, F2, H2, etc.) are the most stable in th...

AtreyiMitra2L

I have two questions regarding this question. 1. How do you know which atoms you need to break/form during this reaction? 2. Where did they get the average energy required to break/form the bonds? Alright, so for the first one. Draw out the lewis structures of each of the reactants and products. On ...

AtreyiMitra2L

This is how I think about it. Imagine work as the direction of the push. If the push is on the system (such as compression as the volume is becoming smaller), it would be work done on the system. Because work is done on the system, we know the work must be done by the surroundings. If the push is on...

AtreyiMitra2L

It would be a closed system. In a closed system, there is a transfer of energy but not of matter. Because it's in a tube, the mercury can't escape. But it can respond by changes in volume based on the transfer of energy through heat. Therefore, its a closed system.

AtreyiMitra2L

So I was working on a problem that had to do with expansion and a subsequent lower internal energy. The question asked if the pressure of the gas would be higher or lower when these changes are completed. I was under the assumption that because there is an inverse relationship between pressure and v...

AtreyiMitra2L

Why is it that with ideal gases, there is no change in internal energy? Thank you!

AtreyiMitra2L

In this question, we are supposed to use bond enthalpies to estimate the reaction enthalpy for 3C2H2(g)-->C6H6(g). In the solutions manual, it says that only 6 C H bonds are formed (where the bond is between a single and double). It seems to me that 4 C - H bonds are formed as well. Can someone plea...

AtreyiMitra2L

i believe it means the way it sounds. the state it is most likely to be found in nature

AtreyiMitra2L

The rest of the question is not important. The question is asking us only to find the change in internal energy. Therefore, we should only be looking for changes in q and w. We find c (the heat capacity) so that we can use this value later to multiply it by the temperture change in the second experi...

AtreyiMitra2L

Today in class we learned that enthalpy is related to heat. Given that temperture is defined as the average kinetic energy of particles, this definition does not make it hard to understand why the two may be correlated.

AtreyiMitra2L

In this question, you have to understand that the heat released by one will be absorbed by the other. Consequently, you will use -q(copper) = q(water). The initial temp of copper is 100 C and the initial temp of water 22 C. Because you know the masses and can find the specific heat, you should autom...

AtreyiMitra2L

How is the volume fixed in the bomb calorimeter and not in the coffee cup calorimeter?

AtreyiMitra2L

yes i am confused by this too. a similar question was on one of the modules 15. The photosynthesis reaction, 6 CO2(g) + 6 H2O(l) ⇌ C6H12O6(aq) + 6 O2(g), is endothermic. What effect will the following changes have on the equilibrium composition. a) Water is added. b) The partial pressure of CO2 is d...

AtreyiMitra2L

on worksheet 9 number 1, the question says that x is the equilbrium concentration of so2. I understand that you have to balance the equation, but I don't understand equilbrium concentration would go from x to 2x when the question said that it should be x. Why is it that is is written as 2x instead?

AtreyiMitra2L

I was working on the 2013 final exam and one of the questions said that the electron configuration for Pd is [Kr]4d10. Can someone explain to me how this is the case?

AtreyiMitra2L

Is ClF3 a lewis acid or base and why?

AtreyiMitra2L

yes this is correct.

AtreyiMitra2L

The formulas are the exact same.

AtreyiMitra2L

The solvent is what dissolves the solute. It is usually a liquid in this class. The universal solvent is water. However, it can also be a gas or supercritical fluid. The solvent can not be aqueous as an aqueous solution means that the solvent was water.

AtreyiMitra2L

Expanded octets become a possibility at 3p. It becomes a possibility bc 3d is right above it. This sub shell can hold more electrons which is why atoms like P, S, and Cl can hold an expanded octet.

AtreyiMitra2L

If the units are already in pressure and it just says k, then you should keep the units in pressure. If the units are already in molarity and it just says k, then you should keep the units in molarity. If it says kc, then make sure to have the units be in molarity. With Kp, have the units be in pres...

AtreyiMitra2L

When it says Kc, you should be finding the concentrations in molarity. When it says k, if the units are pressure, keep it in pressure. If the units are in molarity, keep it in molarity.

AtreyiMitra2L

Hello, I am very confused about the order of the ligands within the coordination sphere. When I asked Dr. Lavelle in class, he said that the order didn't matter. This idea is reaffirmed throughout chemistry community. However, within the textbook, it says that the chemical symbols of the ligands sho...

AtreyiMitra2L

Hello. I am still confused about certain bond angles. I posted this elsewhere but I want to say it again to highlight the discord. This what I feel it to be. Sea-saw : <90, <120, <180 T-structure : <90, <120, <180 Square pyramidal : <90 Square planar : 90 Can someone please verify these all of these...

AtreyiMitra2L

Hi Mike, How water is written does not matter, so long as it is correctly written with two hydrogens and one oxygen. H2O and OH2 are equally correct. For part d., because oxalate has the prefix bis-, this has higher priority than "diaqua" as b precedes d. Hope this helps. But I thought th...

AtreyiMitra2L

First, you would have to change moles to molar concentrations. Then you would put all your initial molar concentrations in your ice chart. Then for change, based on the coefficients of the reactants and products, subtract/add the coefficient x x. You subtract for the reactants and add for the produc...

AtreyiMitra2L

When there are more reactants than products, the equilibrium sits to the left. This is because the denominator of the K expression is larger than the numerator, which leads to a K value that is less than one meaning that the equilibrium sits to the left. Actually, K is actually not affected by the ...

AtreyiMitra2L

Chem_mod said in a different post that they are interchangeable.

AtreyiMitra2L

Remember with K and Q products is over reactants. When Q is less than K, we know that must mean that there are more reactants at this point in the system. If there are more reactants, then the reaction will have the tendency to shift to the products. In terms of Gibbs free energy, the outline says t...

AtreyiMitra2L

Are there any specific exceptions we should be aware of for Chemical Equilibrium (in some other chapters there are specific exceptions to know, like chromium and copper for the orbitals)? I suppose one would be if you yield two positive values from your quadratic, you are two use the value of X tha...

AtreyiMitra2L

Alissa Stanley 3G wrote:So it is possible for Q and K to be equal, if Q is at equilibrium?

Yes! If Q and K are equal, we know that the mixture has equilibrium composition and has no tendency to change in either direction.

AtreyiMitra2L

You determine if its polydentate if there are multiple atoms that have lone pairs on them that can form a coordinate covalent bond. A rule of thumb is that whenever you have Nitrogens, you are sure that the molecule is polydentate. Because there are two nitrogens as there is in ethylenediamine (en) ...

AtreyiMitra2L

Q can be the concentrations of the reactants and products at any point in the reaction. K can only be the concentrations of the reactants at equilibrium. You go about solving them the same way. By using both Q and K, we can determine which way the reaction is more likely to proceed.

AtreyiMitra2L

Well 17.1 was unassigned and unrelated to what we will be tested on so I assume you meant 17.29? Lewis structures aren't needed. This pdf is very helpful with naming compounds. Memorize it. https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14A/NamingCoordinationCompounds.pdf The charges ...

AtreyiMitra2L

RenuChepuru1G wrote:at what point should you figure out cations versus anions and is there an easy way to do that?

The cations are what has an overall positive charge and the anions are what has an overall negative charge.

AtreyiMitra2L

Use this outline as a guideline.

https://lavelle.chem.ucla.edu/wp-conten ... ibrium.pdf

AtreyiMitra2L

Yes, but it would still be a good idea to do practice problems to familiarize yourself with the topics. Sometimes through practice, you learn exceptions that teachers won't directly teach you.

Search found 169 matches