CHEMISTRY COMMUNITY

Kenan Kherallah 2C

705192887 wrote:Can someone who went to the review session post pictures of it? Thanks

Pleaseeee

Kenan Kherallah 2C

On the syllabus it says that we have to know how to derive first second and third orders. I assume that this is because they wont be handed to us on the constants and equations sheet. But I'm a bit worried that we might actually be able to have to derive rate laws from integrated rate laws as I have...

Kenan Kherallah 2C

Can you post the problem please

Kenan Kherallah 2C

I think its more k1 vs k2. In that case the second rate constant given would be k2

Kenan Kherallah 2C

I would also like to know the answer to this question

Kenan Kherallah 2C

Do we have to know how to draw the transition state (question 15.85 on 6th edition) on the final?

Kenan Kherallah 2C

Just to go along with what the person above said, if they tell you the activation energies, the slow step is the one with the higher or larger Ea

Kenan Kherallah 2C

Yup just like 14A it will be cumulative

Kenan Kherallah 2C

Sorry I can't do that

Kenan Kherallah 2C

You would look at the reducing power which is another way of saying how likely the substance is to be oxidized. This means the more negative the E potential the better.

Kenan Kherallah 2C

I would also like to know the answer to this question. What I do know is that Gsys= Wmax

Kenan Kherallah 2C

I dont understand why K = H+/OH-. I thought it should be Kw= H+ times OH-. Can someone explain?

Kenan Kherallah 2C

In cell diagrams in the textbook, they are much more complicated than those that we practiced in class. In many answers, I see the placement of H+ and I just want to know why (worked example 14.7 in the 6th edition shows this). Also, how do we balance this reaction (d) Pt(s) | O2 | H+(aq) || OH-(aq)...

Kenan Kherallah 2C

By finding the gibbs free energy change through summation of products gibbs free energy minus summation of reactants gibbs free energy, we can now set this value equal to = -RT lnK and solve

Kenan Kherallah 2C

Are there notes for Gibbs and electrochemistry?

Kenan Kherallah 2C

In class on Friday Dr. Lavelle said that he was finished covering everything we need for the test 2 but I see that we have not covered anything to do with equilibrium constants in relation to standard electrode potentials (section 14.8 in the 6th edition) with the formula: ln K = nrFE°/RT On his web...

Kenan Kherallah 2C

Expansion relates to an increase in volume occupied (which corrolates to an increase in number of moles)

Kenan Kherallah 2C

Because steam has to go through a phase transition while water doesn't as well as the fact that the enthalpy of condensation is very high, so more energy is released

Kenan Kherallah 2C

I would assume that he wouldn't

Kenan Kherallah 2C

For question 4, Im just wondering if the condition is temperature at zero or if the change (delta T) of temperature is zero so an isothermal condition?

Kenan Kherallah 2C

For the last question, Do we use 0.15 moles from the previous question since I think we have to use moles in the last question

Kenan Kherallah 2C

When we are calculating work with either constant pressure or not, do we always have to use units of atm? Can we use pascals or bar or should we convert?

Kenan Kherallah 2C

I'm not sure but its not as bad as what the textbook makes it look like: for constant volume its 3/2R, 5/2R, or 3R for atoms linear and then nonlinear molecules. For consant pressure its 5/2R, 7/2R, or 4R.

Kenan Kherallah 2C

Jeannine 1I wrote:
005199302 wrote:for #6, shouldn't the work be -12.4kJ?

w=-PxdV = (-122atm/L)(101.33J/atmL) = -12.4kJ

Were you able to solve 6a? I've tried doing it but keep getting 12.85 kJ, but I don't see where it went wrong.

I got the same answer and online I saw that its right so I think you're correct

Kenan Kherallah 2C

Not that I am aware of but I think by the weekend he should have something up

Kenan Kherallah 2C

What is the definition of it and how does it help us

Kenan Kherallah 2C

Calculate the work that must be done against the atmosphere for the expansion of the gaseous products in the combustion of 1.00 mol C6H6(l) at 25 C and 1.00 bar. (b) Using data in Appendix 2A, calculate the standard enthalpy of the reaction. I just need the explanation as to why the answer in part a...

Kenan Kherallah 2C

Make sure to put a negative for the exothermic reaction (this is a common mistake that I usually forget)

Kenan Kherallah 2C

Whats R?

Kenan Kherallah 2C

For question 5, do we have to have whole numbers in the balanced equation to use bond enthalpies or can we keep fractions in there (for this and other questions like this).

Kenan Kherallah 2C

Yes Kw is provided

Kenan Kherallah 2C

Can someone give Lavelle's exact definition of state properties?

Kenan Kherallah 2C

Thank you so much, these really do help

Kenan Kherallah 2C

WIll be tested on multiple protonations or deprotonation events. We haven't talked about it in class but it was included on questions 77 and 81 of the homework.

Kenan Kherallah 2C

the buffer can be made by a weak base. Buffer solutions will not change in pH if there is a small change in the concentration of H+ or OH-. Hence, this means that a buffer solution must have a base and an acid component to react with either an added base or an added acid.

Kenan Kherallah 2C

So just adding on to what Dimitri said, sulfuric acid is a diprotic acid and it is a strong acid for the first dissociation and a weak acid for the second. In the first dissociation the equation is H2SO4 +H20 --> HSO4- + H30+ . this would mean that the 0.15 M of H2SO4 has completely dissociated and ...

Kenan Kherallah 2C

I have two questions: the first is this one. Its included in the homework list of problems but I don't know of any way how to solve this with what we've learned in class. Does someone know if there is a way and if we should know how to do it? Use the information in Table 11.2 to determine the value ...

Kenan Kherallah 2C

I dont think we will be using this equation since he hasnt mentioned it in lecture and all HW problems are solvable without the equation.

Kenan Kherallah 2C

If they specify that they want Kp then you should write equilibrium constant in terms of partial pressures of the reactants and products. However, if you have substances that are not in the gaseous phase then you can't use Kp and must use Kc. If the book or the question doesn't specify whether Kc or...

Kenan Kherallah 2C

11.79 A reactor for the production of ammonia by the Haber process is found to be at equilibrium with PN2 3.11 bar, PH2 1.64 bar, and PNH3 23.72 bar. If the partial pressure of N2 is increased by 1.57 bar, what will be the partial pressure of each gas once equilibrium is re-established? I have no id...

Kenan Kherallah 2C

11.7 The following flasks show the dissociation of a diatomic molecule, X2, over time. (a) Which flask represents the point in time at which the reaction has reached equilibrium? (b) What percentage of the X2 molecules have decomposed at equilibrium? (c) Assuming that the initial pressure of X2 was ...

Kenan Kherallah 2C

Does anyone know how to do 2B please on ws 9

Kenan Kherallah 2C

Hey guys can someone help me with question 21, I really dont know how to approach this since I just get a formal charge of +2 in total for the ligands which doesnt seem right

Kenan Kherallah 2C

Im just wondering why the tail has a -1 charge since for me when I calculated the formal charges of the nitrogens connected to Co I got +1, +1, +1, +1, and 0
Wouldnt that give an overall +4 charge for the tail?

Kenan Kherallah 2C

No and this is also why for vsepr any resonance structure can be used

Kenan Kherallah 2C

It would be trigonal pyramidal since there is a lone pairs of electrons on the central atom in a tetrahedral arrangemen

Kenan Kherallah 2C

Lone pairs repwl bonding pairs more than bonding pairs repel each other causing the bond angles to be smaller than usual.

Kenan Kherallah 2C

So sp1 has a 180 degree angle, sp2 has a 120 angle and sp3 has a 109.5 angle so as the s character decreases the angle will smaller. Therefore if it increases rhen the angle increases

Kenan Kherallah 2C

That's because they have a triple bond between them so they have 2 electron domains= sp1

Kenan Kherallah 2C

The two strands of the nucleic acid DNA are held together by hydrogen bonding between four organic bases. The structure of one of these bases, thymine, is shown below. (a) How many protons can this base accept? (b) Draw the structure of each conjugate acid that can be formed. (c) Mark with an asteri...

Kenan Kherallah 2C

So for Test #3, my TA started drawing sigma and pi bonds but I remember that Lavelle said that we don't need to know how to do this, but my TA said that we do in fact need to know how to do this for test #3.

Is this true?

Kenan Kherallah 2C

So N3- is more polarizable than O2- because N3- is actually larger than O2- as well as less electronegative. My TA said that the trends for polarizability is actually the larger the atom is and the less electronegative, the more polarizable it is. So I dont think the charge here matters, its just te...

Kenan Kherallah 2C

I think you are referring to the dipole moment. I asked my TA this and she said she doesn't think so but it doesn't hurt to know it, especially for individual bond dipoles. For this, you just put the arrow to the more electronegative atom.

Kenan Kherallah 2C

Yes we do not need to memorize only know that AX4 is a tetrahedral and AX4E is trigonal pyramidal, etc

Kenan Kherallah 2C

Can someone explain how to draw KrF4

Kenan Kherallah 2C

The energy of the attractive forces

Kenan Kherallah 2C

VSEPR also gives you structures based on experimentally observed results. Hence, the rules outlined in VSEPR will allow you to essentially represent structures in a 2-D manner.

Kenan Kherallah 2C

Guys do you know if for the upcoming test if it is after or before thanksgiving. If its before, what happens if we dont have our discussion on that day due to break.

Kenan Kherallah 2C

I didnt but I wrote down that pentane was a mobile fluid, pentadecane was a viscous fluid and octadecane was a waxy fluid. I dont know if that helps though.

Kenan Kherallah 2C

The Lewis base is an electron pair donor and so it donates it's lonely pair to a Lewis acid which usually has an emptru orbital to take those electrons. So the Lewis acid is a electron pair acceptor and a dative bond is formed.

Kenan Kherallah 2C

Just adding to what the person above said. Those exceptions also work for the elements below Cr and Cu (directly below).

Hope this helps!

Kenan Kherallah 2C

Heyy I was wondering where I can find the answers for worksheet 5 (newest one)? Also for question 9 aren't there more than 2 ions that aren't isoelectronic with F-. I got 4 ions (Li+, Be2+,O-, N2-) that aren't isoelectronic.

Thanks!

Kenan Kherallah 2C

Thank you so much for the extra practice!

Kenan Kherallah 2C

Does anyone know how to do question 13 a and b? I though I could do 13a but thirteen b said that what she did (which is what I did) was wrong.

Kenan Kherallah 2C

I think it will also cover everything we've been doing up to the midterm (fundamentals plus chapters 1 2 and 3) :).

Kenan Kherallah 2C

to get the number of orbitals you use the formula: number of orbitals= nsquared (n being the principal quantum number). To get the number of electrons you just multiply nsquared by 2.

Kenan Kherallah 2C

Im not sure about the first part of teh question but i can say that SI units are needed so meters per second should be used!

Kenan Kherallah 2C

I believe there are few metallic surfaces that would emit electrons if visible light is used but the majority only get emitted if ultraviolet light is used

Kenan Kherallah 2C

Well since the question said that lines in the balmer series are observed, we can know that n1= 2 since the balmer series has a lower energy level of n=2 (visible light region). Furthermore, knowing this we can assume that 410.2 nm corresponds to n=3, 434 nm: corresponds to n=4, and so on. The next ...

Kenan Kherallah 2C

I'm still confused on how you were able to identify that those wavelengths correspond to those lines. Is there a table I should be looking at?

Kenan Kherallah 2C

You also use p = h times wavelength ^-1 because this is the only particular way for us to get the desired units

Kenan Kherallah 2C

You could also kind of gage at the question as well. Large objects like a car or a lamp wont have a noticeable wavelength but at the atomic scale, electrons and neutrons and such will.

Kenan Kherallah 2C

Use the Rydberg formula for atomic hydrogen to calculate the wavelength of radiation generated by the transition from n = 4 to n = 2. (b) What is the name given to the spectroscopic series to which this transition belongs? (c) Use Table 1.1 to determine the region of the spectrum in which the transi...

Kenan Kherallah 2C

Hello Jennifer, I think that if the frequency increases the extent of the change in the electric field also increases since increasing the frequency means that the radiation has an increase in energy (more oscillation and interaction with electrons). The opposite can be said if the frequency decreas...

Kenan Kherallah 2C

I need help with the following question: Lines in the Balmer series of the hydrogen spectrum are observed at 656.3, 486.1, 434.0, and 410.2 nm. What is the wavelength of the next line in the series?

I get that n1=2 but what is n2? Thanks!

Kenan Kherallah 2C

In our last class, Dr. Lavelle said that it is required.

Kenan Kherallah 2C

I think explaining it would be better through examples. 1.0 has two sig figs 10 has one sig fig, so does 100, so does 2000. 2300 has two sig figs (so numbers before zeros count as sig figs). But if a decimal is placed then the zeros are significant e.g 2300. has 4 sig figs. 0.003 has one sig fig 0.0...

Kenan Kherallah 2C

Hello,

So I would start by getting the moles of Na2CO3 needed. You would do this by 0.15 * 0.0234 M to get 0.00351.
By doing this you then have to solve for the volume of the concentration 0.778M. 0.00351/0.778 = 0.00451 L

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