Search found 20 matches
- Tue Mar 13, 2018 12:47 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: 15.65 c
- Replies: 2
- Views: 211
15.65 c
15.65 For the reversible, one-step reaction 2A ⇌ B + C, the forward rate constant for the formation of B is 265 L.mol^-1.min^-1 and the rate constant for the reverse reaction is 392 L.mol^-1.min^-1. The activation energy for the forward reaction is 39.7 kJ.mol^-1 and that of the reverse reaction is ...
- Tue Mar 13, 2018 12:41 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: 9.23
- Replies: 4
- Views: 419
Re: 9.23
The positional disorder in COF2 is greater than in BF3 because BF3 can only be arranged in one way--with B as the central atom connected to 3 F atoms. However, COF2 can be arranged in three different ways, giving it a higher molar entropy than BF3 as can be seen through the equation S = k ln(W)
- Tue Mar 13, 2018 12:34 pm
- Forum: Phase Changes & Related Calculations
- Topic: Test #1 question
- Replies: 2
- Views: 216
Re: Test #1 question
Since the system is isothermal, we know that the change in internal energy is 0. So q+w=0, and q=-w. Since the balloon is expanding its volume against a changing pressure, work is being done. This also means that the energy lost during work is gained as heat, because the internal energy must stay 0....
- Sun Feb 18, 2018 9:52 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.15A
- Replies: 4
- Views: 303
Re: 14.15A
There is no 2Br in the equation
- Sun Feb 18, 2018 9:49 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.17A
- Replies: 1
- Views: 181
Re: 14.17A
No, all the information needed from appendex b would be provided if this question was on a test or exam.
- Sun Feb 18, 2018 9:47 pm
- Forum: Balancing Redox Reactions
- Topic: 14.1
- Replies: 5
- Views: 347
14.1
I am confused as to how the chemical equation on 14.1, Part D was balanced.
- Sun Feb 11, 2018 6:46 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: 11.5
- Replies: 1
- Views: 153
Re: 11.5
I don't think number 55 is part of what is covered on the midterm, but to solve it, you have to use the ice box method
- Sun Feb 11, 2018 6:43 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Practice Midterm
- Replies: 1
- Views: 213
Re: Practice Midterm
It is under the "Administrative Questions and Class Announcements" topic. Here is the link
viewtopic.php?f=160&t=27112
viewtopic.php?f=160&t=27112
- Sun Feb 11, 2018 6:40 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Derivation of Formulas
- Replies: 5
- Views: 505
Re: Derivation of Formulas
No, we never need to know the derivations, they are just used to help us understand the formulas
- Sun Feb 04, 2018 7:55 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: 9.35
- Replies: 2
- Views: 176
9.35
Why is container A's entropy is higher than container C's entropy if Container A's atoms are monatomic while C's atoms are diatomic? If they're diatomic, doesn't it make it a bit more complex?
- Sun Feb 04, 2018 7:36 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 8.53
- Replies: 2
- Views: 165
8.53
I don't really understand how to set up number 53, how do I set up the balanced equation?
- Sun Feb 04, 2018 7:32 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: 9.33
- Replies: 1
- Views: 156
Re: 9.33
Yes, this is true for gaseous products. However, if there are a different amount of moles in a different phase, the answer could be different
- Sun Jan 21, 2018 9:48 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Bomb calorimeter
- Replies: 7
- Views: 459
Re: Bomb calorimeter
It's an isolated system because no energy or matter is exchanged between the calorimeter and the surroundings.
- Sun Jan 21, 2018 9:27 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.75 part b
- Replies: 3
- Views: 237
Re: 8.75 part b
A C-C bond is formed because when the double bond C=C is broken, the C-C bond has to be reformed. It is not possible for one of the C=C bonds to be broken only. You are right about the C-H bond, but the other bond that forms is C-OH, which is not a double bond.
- Sun Jan 21, 2018 9:21 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.67
- Replies: 3
- Views: 279
Re: 8.67
I did this question by using the bond enthalpies. The total enthalpy required to break the bonds was 684 kj.mol^-1 but the total enthalpy of the formation of the bonds was -926 kj.mol^-1 and is negative since when bonds form, it is an exothermic process. Add -926 to 684 and you get -242 kj.mol^-1
- Fri Jan 12, 2018 2:19 pm
- Forum: Phase Changes & Related Calculations
- Topic: qv vs qp
- Replies: 4
- Views: 1205
Re: qv vs qp
Qp is constant pressure while Qv is constant volume
- Sun Oct 15, 2017 12:14 pm
- Forum: DeBroglie Equation
- Topic: 1.55
- Replies: 2
- Views: 219
1.55
In question number 55, the question states that "absorption takes place at the infared spectrum at 3600cm^-1. Is this referring to the wavelength, or the diameter?
- Sun Oct 15, 2017 11:54 am
- Forum: DeBroglie Equation
- Topic: Post-Module Assessment #35 [ENDORSED]
- Replies: 2
- Views: 398
Re: Post-Module Assessment #35 [ENDORSED]
it doesn't have wavelike properties because it is too small to detect. the threshold is anything x 10^-15 as the smallest wavelength that portrays wavelike properties
- Fri Oct 06, 2017 3:08 pm
- Forum: SI Units, Unit Conversions
- Topic: SI units [ENDORSED]
- Replies: 6
- Views: 577
Re: SI units [ENDORSED]
Also not that there are 1000 milimoles in a single mole. I saw this appear a few times throughout the homework. Try to remember the prefixes as it'll help a lot.
- Thu Oct 05, 2017 8:00 pm
- Forum: Empirical & Molecular Formulas
- Topic: Fundamental F, F3
- Replies: 1
- Views: 251
Re: Fundamental F, F3
Since oxygen has the highest molar mass and occurs the most (there is only one atom of hydrogen and one atom of nitrogen but there are three oxygen atoms in the molecule), you can assume that the element that occurs with the greatest mass percentage is oxygen.