Search found 12 matches
- Mon Mar 13, 2017 3:06 pm
- Forum: *Ketones
- Topic: Ketone Naming
- Replies: 1
- Views: 1269
Re: Ketone Naming
The best way to understand this would probably be to draw out the structures. You will notice that up until there is 5 carbons, the location is obvious simply by the name butane or propanone. With propanone, only one carbon is available for the oxygen to be on because if it was on either of the end ...
- Sun Mar 12, 2017 3:30 pm
- Forum: *Constitutional and Geometric Isomers (cis, Z and trans, E)
- Topic: Cyclo-isomers
- Replies: 4
- Views: 856
Re: Cyclo-isomers
I could be wrong but I believe I drew cyclopropane with an ethyl substituent, cyclobutane with a methyl substituent, and cyclopentane
- Sun Mar 12, 2017 3:26 pm
- Forum: *Aldehydes
- Topic: trans and cis attachments to cyclic carbon structures
- Replies: 1
- Views: 1340
Re: trans and cis attachments to cyclic carbon structures
I found this a little confusing as well, but understood it more after I thought of the benzene ring as a substituent. This benzene ring substituent is trans to carbon three, where O is attached to the double bond.
- Thu Mar 09, 2017 3:46 pm
- Forum: *Alkanes
- Topic: Naming Help
- Replies: 8
- Views: 2192
Re: Naming Help
Also, if you have more than two substituents you start with the substituents that are closest together. For example you would put 1,2 and 4 over putting 1 and 3,4
- Thu Mar 09, 2017 3:42 pm
- Forum: *Free Energy of Activation vs Activation Energy
- Topic: HW #4.40
- Replies: 2
- Views: 1470
Re: HW #4.40
you can also simply assume, as a common theme, that any activation over 80kj/mol and above will not take place at room temperature.
- Mon Mar 06, 2017 5:49 pm
- Forum: *Electrophilic Addition
- Topic: Transition step
- Replies: 3
- Views: 1602
Re: Transition step
yes the first step is always slower because it involves breaking a bond and forming a high energy intermediate whereas in the second it only has to break that intermediate and form a new lower energy bond.
- Wed Feb 15, 2017 7:48 am
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Nernst Equation with ln vs. log
- Replies: 2
- Views: 2315
Re: Nernst Equation with ln vs. log
the formula with ln multiplied by 2.303 will give you that value. moles transferred and given should be equivalent because you should have balanced your anode and cathode equations so that electron amounts are the same and can be cancelled.
- Fri Feb 10, 2017 12:29 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Ch. 14 HW #9a
- Replies: 1
- Views: 486
Re: Ch. 14 HW #9a
Because it's ΔGr° instead of ΔG°, the units are joules per mole. Since nr is in molar form then it is just a plain number and has no units to it meaning your final answer will have moles in it because there was no mole unit to cancel out the mole units in faraday's constant
- Wed Feb 01, 2017 12:42 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Free energy at equilibrium
- Replies: 2
- Views: 532
Re: Free energy at equilibrium
The standard free energy change is a constant for a reaction at a given pressure and temperature.The free energy change (NOT standard) is the total free energy changes of all the components in the reaction at any given point. The standard free energy change is constant throughout the reaction while ...
- Tue Jan 24, 2017 12:53 pm
- Forum: Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy
- Topic: Why are W and V interchangeable in the Boltzman formula?
- Replies: 2
- Views: 441
Re: Why are W and V interchangeable in the Boltzman formula?
I don't know that they will always be interchangeable but in the example given during lecture the degeneracy (W) was interchangeable with volume because he said that the volume doubled. Doubling the volume therefore making 2x the available positions. This means that V2/V1=2 and W2/W1 are both equal ...
- Thu Jan 19, 2017 4:39 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Work equation
- Replies: 1
- Views: 359
Re: Work equation
The summation integrates all of the infinitesimal amounts of work done as the volume changes from initial to final. Essentially, you find your total amount of work by adding together all the tiny amounts of work done in each step as the volume progressively changes. w=-P(deltaV) for irreversible equ...
- Thu Jan 19, 2017 4:27 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: First Law of Thermodynamics Practice Problem [ENDORSED]
- Replies: 1
- Views: 577
Re: First Law of Thermodynamics Practice Problem [ENDORSED]
yes, when you get a negative answer the system is doing work and if it is positive then the system had work done on it.