Search found 12 matches
- Sat Mar 12, 2016 12:37 pm
- Forum: *Alkanes and Substituted Alkanes (Staggered, Eclipsed, Gauche, Anti, Newman Projections)
- Topic: Final Question
- Replies: 2
- Views: 608
Final Question
Are we expected to memorize the energies for the different conformations (for ex. gauche= .9kcal/mol) and also the c-c-c- angles cyclopropane, cyclobutane, and cyclohexane?
- Sat Mar 12, 2016 12:29 pm
- Forum: *Alkanes and Substituted Alkanes (Staggered, Eclipsed, Gauche, Anti, Newman Projections)
- Topic: Ch3 in Newman projection
- Replies: 1
- Views: 531
Ch3 in Newman projection
When looking down a bond such as 2-3 bond for butane, does the CH3 in the front and back refer to carbon 1 and carbon 4?
- Tue Mar 08, 2016 7:37 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 2989285
Re: Chemistry Jokes
What's the name of an non hydrogen atom orbital that has a talk show?
Ellen Degeneracy!
Ellen Degeneracy!
- Thu Feb 25, 2016 1:16 pm
- Forum: *Alkanes
- Topic: Writing Out Condensed Structural Formulas
- Replies: 1
- Views: 349
Writing Out Condensed Structural Formulas
When writing condensed structural formulas, does it matter the order of the placement of the constituent or is it conventionally put at the end? For example in Self Test 1.1 A, the condensed version is written as (CH3)2CHCH2CH3
- Mon Feb 15, 2016 2:29 pm
- Forum: General Rate Laws
- Topic: Using M instead of molL-
- Replies: 1
- Views: 585
Using M instead of molL-
In the textbook solutions, all answers are in molL-. Is it fine if we just use M?
- Mon Feb 15, 2016 2:11 pm
- Forum: General Rate Laws
- Topic: Including k in rate laws
- Replies: 4
- Views: 897
Including k in rate laws
When asked to write down the rate law of a reaction, do we need to solve for k? Or are we just expected to write "k" for the rate law?
- Tue Feb 09, 2016 8:30 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: W=0 when number of gas molecules constant
- Replies: 1
- Views: 397
W=0 when number of gas molecules constant
In Midterm 2015 Q2&3 Part A the answers state that w=0 because the product and reactant both have 2 moles gas molecules (V=0). So for any circumstances where P=0, would you just have to compare the number of moles of gas molecules on each side of the equation to know that volume is constant, ign...
- Mon Feb 08, 2016 6:09 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Midterm 2011 Q5
- Replies: 1
- Views: 421
Midterm 2011 Q5
Why in Part A is the answer in JK-1mol-1 but the answer in Part b is in kJ (lacking the "mol-1"). I would understand why the mol-1 would be left out because both questions are asking specifically for one mol. So why is there still the "mol-1" for the answer in part A?
- Mon Feb 08, 2016 3:46 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 2011 Midterm Q1B
- Replies: 1
- Views: 409
2011 Midterm Q1B
Can someone explain conceptually why the the decrease in internal energy results in a decrease in temperature and pressure?
- Tue Feb 02, 2016 9:05 pm
- Forum: Balancing Redox Reactions
- Topic: Balancing Basic Redox Reactions (HW 14.5 A)
- Replies: 2
- Views: 712
Balancing Basic Redox Reactions (HW 14.5 A)
For this problem, how do you know to split up the redox reactions into O3->02 and Br->BrO3, especially because in 14.3 D you put the reactant(Cl2) in both half reactions?
- Sat Jan 30, 2016 9:21 pm
- Forum: Balancing Redox Reactions
- Topic: oxidation and reduction
- Replies: 2
- Views: 543
Re: oxidation and reduction
What is being oxidized (losing electrons) is the reducing agent. What is being reduced (gaining e-1) is the oxidizing agent.
- Mon Jan 18, 2016 4:03 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Heat Release Quantities
- Replies: 1
- Views: 503
Re: Heat Release Quantities
Since enthalpy density is defined as enthalpy released per liter, the value would be positive since heat is being released in this case. If it were defined as enthalpy absorbed in the reaction per liter, the sign would be negative, as you calculated.