CHEMISTRY COMMUNITY

Sarah_Heesacker_3B

You use cis and trans (or the E,Z naming system) if there is a double or triple bond. When there are single bonds, they can rotate freely so there is no different name for the structures when they rotate. However, when there are pi bonds like in double bonds, they cannot rotate. This gives two diffe...

Sarah_Heesacker_3B

You determine which name has lower numbers and choose that one. Since 2 is less than 3, you already know that the first name is the correct one.

Sarah_Heesacker_3B

I think that the reaction profile would be the same, but the labels would be different. Step 1 would have a larger activation energy because the activation energy for Step 1 is measured from the reactants to the first transition state (from start of graph to top of first peak). Activation energy for...

Sarah_Heesacker_3B

I believe the green Organic Chemistry book is the textbook with text, homework problems, solutions. In class we should still only need the blue course readers that have the lecture notes.

Sarah_Heesacker_3B

The order also reveals how many molecules are involved. For example, a second order reaction involves two molecules colliding at the same time. This is why reactions of order 3 or more are uncommon because it is unlikely for 3 or more molecules to collide at the exact same time.

Sarah_Heesacker_3B

In that question, n is actually equal to 2. The half reactions are Fe2+(aq) +2e- --> Fe(s) and Ce4+(aq) + 1e- --> Ce3+(aq). So when the equations are balanced the number of electrons in each equation is 2, which is why n=2.

Sarah_Heesacker_3B

It is possible, but as Dr. Lavelle mentioned today it is very rare because a 3rd or 4th order reaction, for example, would require 3 or 4 molecules all to collide at the exact same time. This is why the 1st and 2nd order reactions are the most common.

Sarah_Heesacker_3B

Yes, you raise the number of positions to the power of the number of particles. You use Avogadro's number when it says that you have 1 mol. However, you do not raise W to Avogadro's number, but use Avogadro's number as the number of particles (so you raise the number of positions/microstates to Avog...

Sarah_Heesacker_3B

I think it means graphite, because elemental carbon can appear in multiple forms, such as graphite and diamond.

Sarah_Heesacker_3B

These values are given on the equation sheet on the back of the periodic table we received with the course reader. I think this means that they will likely be given to us on any tests or quizzes, so I don't think we need to memorize them.

Sarah_Heesacker_3B

I believe so! You can check by comparing the values of the bond enthalpies used in the example in the course reader to the values in Table 8.7 in the textbook. Based on these values used, the C=C double bond does break completely (612 kJ/mol) and then a C-C single bond forms (-348 kJ/mol). Hope this...

Sarah_Heesacker_3B

Since the problem asked you to find the Ka values (which can be found by taking the antilog of the pKa value given), you can use those to determine the strength of the acid. A larger Ka value means the acid is stronger. Since H3PO3 has a Ka value of .01 and H3PO4 has a Ka value of .0076, H3PO3 has a...

Sarah_Heesacker_3B

Also, the reason it is from 0 to 14 is because pKa + pKb = pKw. pKw is -log(Kw)= -log(10^-14)= 14.

Sarah_Heesacker_3B

You have to look at the charges of the other molecules in the compound. In that example, there is Cl3 which would have a charge of -3 because each Cl has a charge of-1. H20 is neutral and therefore does not have any charge. The negative outside of the bracket indicates that the total charge of the c...

Sarah_Heesacker_3B

I think that the MO diagrams only include the valence electrons, so for most of the examples we've done like oxygen, the 1s orbital would not be included.

Sarah_Heesacker_3B

Also, the problem specifies that HClO3 is an oxoacid, which means that the hydrogen is bonded to the oxygen and not the central atom, which in this case is chlorine.

Sarah_Heesacker_3B

Yes atomic radius does affect bond lengths because if there is a larger distance between atoms, the bond is longer and therefore weaker. Triple bonds are the shortest bonds because the atoms are sharing more electrons, which is why a triple bond would be stronger than a double bond which is stronger...

Sarah_Heesacker_3B

Madeline, I was confused about that at first as well, but I think that they must follow Hund's Rule to be in the ground state. Since a, b, and c don't follow Hund's Rule, they are in the excited state. Applying Hund's Rule gives the lowest energy electron configuration, which is the ground state, so...

Sarah_Heesacker_3B

Yes, all combustion reactions usually occur in the presence of oxygen and produce carbon dioxide and water.

Sarah_Heesacker_3B

Thank you for posting the video; it does clarify the differences some. However, I did not learn about orbitals in high school so I am still slightly confused. I know Dr. Lavelle said that what we went over in lecture were just mathematical models, but I am still not sure how that relates to actual s...

Sarah_Heesacker_3B

If you have not already, watching the module on Limiting Reactant Calculations is extremely helpful and gives these steps to solving these types of problems: 1. Identify reactants and products (In your example, the reactants are Boron Oxide and Magnesium. The products are Boron and Magnesium Oxide.)...

CHEMISTRY COMMUNITY

Search found 21 matches

Re: When to use cis or trans?

Re: Different Substituent Numbers

Re: Self Test 4.4 A

Re: Organic Chem Course Reader

Re: Order Reactions [ENDORSED]

Re: Winter 2013 Midterm 8B

Re: Reaction order

Re: Using the Boltzmann Equation [ENDORSED]

Re: Phase subscript question

Re: Question 8.31

Re: Bonding Enthalpies

Re: HW problem 12.35

Re: pKa and pKb

Re: How to find the roman numeral when naming compounds [ENDORSED]

Re: MO Diagrams

Re: Fall 2014 Midterm #7

Re: Determining Which Has A Stronger Bond

Re: Chapter 2 HW: #39 [ENDORSED]

Re: Balanced Equation for combustion of CH3OH

Re: Shells, Subshells, and Orbitals

Re: Determining Limiting Reactants and Products produced. [ENDORSED]