Search found 12 matches
- Thu Mar 16, 2017 9:32 pm
- Forum: *Alkanes and Substituted Alkanes (Staggered, Eclipsed, Gauche, Anti, Newman Projections)
- Topic: self test 3.3A
- Replies: 1
- Views: 318
self test 3.3A
why would there be more molecules in the gauche conformation at higher temperatures?
- Tue Mar 07, 2017 6:03 pm
- Forum: *Organic Reaction Mechanisms in General
- Topic: 4.26 ochem textbook
- Replies: 3
- Views: 598
4.26 ochem textbook
In this question it asks to complete the reaction mechanism, and in the answer the Iodine is on the left carbon and the Hydrogen is on the right carbon, but can it be the other way around? (hydrogen on left carbon, iodine on right carbon)
- Tue Feb 28, 2017 7:49 pm
- Forum: *Electrophiles
- Topic: 4.11, 4.13
- Replies: 2
- Views: 487
4.11, 4.13
At this point are we supposed to be able to give an example of an Organolithium compound? or an alkyl Grignard reactant? and give examples of alkenes and alkynes? It asks these kinds of questions throughout the section but there isn't very much information about them in the text
- Tue Feb 14, 2017 7:25 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 9.55
- Replies: 1
- Views: 305
9.55
It asks to write balanced equations before calculating enthalpy, entropy, and gibbs free energy. Does the equation need to have the smallest possible molar ratios? I balanced the equation as 3H2 (g) + N2 (g) --> 2NH3 (g) and I got a different answer for entropy than the solutions manual
- Mon Feb 13, 2017 6:38 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.11 d)
- Replies: 2
- Views: 376
14.11 d)
So for the right side of the cell diagram the half-reaction should be O2 (g) + 2H2O (l) + 4e- --> 4OH - (aq) . **originally just O2 (g) --> OH- (aq)** I don't understand why the water molecules are on the reactants side if the first step to balancing the equation is to balance the oxygen molecules (...
- Thu Feb 09, 2017 5:45 pm
- Forum: Balancing Redox Reactions
- Topic: 14.3
- Replies: 2
- Views: 352
Re: 14.3
thanks!
- Thu Feb 09, 2017 1:47 pm
- Forum: Balancing Redox Reactions
- Topic: 14.3
- Replies: 2
- Views: 352
14.3
the skeletal equation is: Cl2 (g) + S2O3 2- (aq) --> Cl- (aq) + SO4 2- Chorine gas is gaining just 1 electron... okay so in the solutions manual the reduction half-reaction is Cl2 (g) + 2e- --> 2Cl- (aq) and I do not understand why we add 2 electrons; that would make the left side have a charge of 2...
- Mon Jan 30, 2017 1:21 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 9.7
- Replies: 1
- Views: 254
9.7
"Assuming that the heat capacity of a ideal gas is independent of temperature, calculate the entropy change associated with raising the temp 1.00 mol of ideal gas atoms.... "
How do we know that this is a monatomic gas?
How do we know that this is a monatomic gas?
- Sun Jan 29, 2017 4:21 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Question 8.45
- Replies: 1
- Views: 295
Question 8.45
The units given for delta H of the reaction is given as 358.8 kJ (not per mole), but the solutions manual gives the enthalpy in kJ/mol and so delta H is multiplied by the 1.25 mols of Sulfur in order to cancel the units. Why did they just change the units? Anyways I thought kJ/mol was used for phase...
- Sun Jan 22, 2017 6:49 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: HW 8.67
- Replies: 3
- Views: 589
Re: HW 8.67
Graphite is solid, but we need it to be in the form of a gas. Sublimation is Hvapor - Hsolid, and the problem said the sublimation of carbon is +717 kJ*mol, so 717- 0 (enthalpy of carbon in it's most stable form graphite is 0) = +717kJ*mol^-1.
- Fri Jan 20, 2017 2:19 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: HW 8.67 b) [ENDORSED]
- Replies: 1
- Views: 301
HW 8.67 b) [ENDORSED]
I thought that the formation of methanol was CO + 2H2. While it makes sense to me why it is actually set up C (gr) + 2H2 (g) + 1/2O2, I don't understand how we are supposed to know that this is the way it's done for this particular problem. CO is bonded by a triple bond, which is a mean bond enthalp...
- Mon Jan 16, 2017 11:16 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Problem 8.57 (Hess's Law)
- Replies: 2
- Views: 434
Problem 8.57 (Hess's Law)
When writing out the balanced equations for the reaction given do we need to use the most simplified (in terms of the moles) equation? For example, for the combustion of H2 I have it written as 2H2 + O2 --> 2H2O but the solutions manual has it more simplified (H2 + 1/2O2 --> H2O), and so the answers...