Search found 10 matches
- Sat Mar 18, 2017 8:27 pm
- Forum: *Alkanes
- Topic: Naming [ENDORSED]
- Replies: 93
- Views: 16715
Re: Naming [ENDORSED]
If it were something like 1-bromooctane I believe you still can write it like that or you can write 1-bromo-octane. Same would apply to 1-bromoethane. I dont think vowels change the naming of the organic molecule
- Sun Mar 12, 2017 7:18 pm
- Forum: *Cycloalkanes
- Topic: Cyclo-
- Replies: 7
- Views: 2300
Re: Cyclo-
Also, I believe in lecture it was mentioned that if there is a double or triple bond in the structure, we start naming from the carbon on the double or triple bond then count the substituents
- Sun Mar 05, 2017 6:51 pm
- Forum: *Alkanes
- Topic: Naming Organic Compounds
- Replies: 4
- Views: 830
Re: Naming Organic Compounds
This actually doesn't work, if we read the textbook it specifically brings up that thought. On page 8 it says "2,7,8-trimethyldecane is the correct name even though the sum of the numbers (2+7+8=17) is higher than the sum, 16, obtained from the incorrect name 3,4,9-trimethyldecane." So we ...
- Sun Feb 26, 2017 11:11 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Slow Reaction
- Replies: 11
- Views: 1981
Re: Slow Reaction
When it comes to the slow step reaction is the reaction that is reversible also the slow step reaction? So if it states that the reaction is reversible k and k(prime) would be included in the slow step reaction as well?
- Wed Feb 15, 2017 11:08 am
- Forum: Administrative Questions and Class Announcements
- Topic: Midterm Winter 2017
- Replies: 87
- Views: 21120
Re: Midterm Winter 2017
Will values like specific heat of different compounds/ elements be given when they're necessary or should I memorize a few basic ones like water, ice and steam? It was said we would be given equations and values that we would need for the test, so the specific heat, the value of R and F etc should ...
- Sun Feb 12, 2017 2:23 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Calculating K for reaction using half reactions [ENDORSED]
- Replies: 4
- Views: 1020
Re: Calculating K for reaction using half reactions [ENDORSED]
In the explanation for the question it says that we should expect E to be negative because the K to be very small, and in order of that to happen the the value of E(naught) must be negative
- Sun Feb 05, 2017 5:10 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Salt Bridge
- Replies: 6
- Views: 1952
Re: Salt Bridge
The metal acts as the conductor so that the electrons can move, so I wouldn't believe the metal ions would be the one that is being transferred as the electrons are moving from areas of high concentration to areas of low concentration causing the charge of the two substances to increase or decrease....
- Sun Jan 29, 2017 11:27 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Equilibrium Constants
- Replies: 1
- Views: 499
Re: Equilibrium Constants
Because the equation is that delta G*=-RTlnK and we know K= [P]/[R] So if the K=1 delta G would equal -RTln1 and ln1=0 If K<1 (when R is greater than P) then it would be ln(.5 for example) which is a negative number. The natural log of all numbers less than 1 and greater than 0 is negative. Which wo...
- Sun Jan 22, 2017 3:00 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Unit Question [ENDORSED]
- Replies: 4
- Views: 970
Re: Unit Question [ENDORSED]
The equation for work is work=P*deltaV and the units that come from that is
P*deltaV= kg*m^-1*s^-2(m^3)= kg*m^2*s^-2 which equates to Joules
So if im not mistaken, the units for work should come out in Joules
P*deltaV= kg*m^-1*s^-2(m^3)= kg*m^2*s^-2 which equates to Joules
So if im not mistaken, the units for work should come out in Joules
- Sat Jan 14, 2017 2:22 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: heat q(sys)+ q(surr) =0
- Replies: 3
- Views: 1250
Re: heat q(sys)+ q(surr) =0
I believe it would still apply, because the heat that is being released into the aqueous system will still be moved into the surrounding of it. So we define our system as the closed system with an aqueous solution that you described and when the heat energy is released it is being released into the ...