Search found 26 matches
- Sat Mar 18, 2017 10:25 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Course Reader Final Practice "2016 Final" Q4 part B
- Replies: 5
- Views: 1189
Re: Course Reader Final Practice "2016 Final" Q4 part B
You can see that A does not work because the intermediates do not cancel out the cumulative reactants for B should be 2CLO and O2 since one of the O2's are cancelled out So the two valid ones are B&C Student B: 2ClO; O2 Student C: 2ClO; O2 you can tell that the second steps are the rate limititi...
- Sat Mar 18, 2017 1:10 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Winter 2013 Final Q4B
- Replies: 1
- Views: 489
Re: Winter 2013 Final Q4B
[Current] = [Initial] + [Change in initial] [O2] = [O2 initial] + [change in initial O2] Flip both equations: 1 / [O2] = 1 / ([O2 initial] + [change in initial O2]) Since Q = [Products] / [Reactants] = 1 / [O2], plug in: Q = 942.23 = 1 / ([.00117] + [Change in initial O2]) Rearrange and solve using ...
- Sat Mar 18, 2017 1:06 pm
- Forum: General Science Questions
- Topic: Winter Final 2013 Q4
- Replies: 1
- Views: 775
Re: Winter Final 2013 Q4
n is based on the total number of electrons transferred in the redox reaction. Fe2+ goes to Fe, meaning that two electrons are transferred per Fe, and there's a molar coefficient of 2, so in the net reaction we transfer 4 electrons.
- Sat Mar 18, 2017 12:58 pm
- Forum: *Electrophiles
- Topic: Is carbon monoxide an electrophile
- Replies: 1
- Views: 2632
Re: Is carbon monoxide an electrophile
Carbon Monixide is an Electorphile. The C=O bond is very polar and the carbonyl carbon is very positive. The carbonyl bond is very polar. There is a partial positive charge on the carbon and a partial negative charge on the oxygen, because oxygen is more electronegative than carbon. This charge sepa...
- Sat Mar 18, 2017 12:53 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Ch 14 #107
- Replies: 1
- Views: 619
Re: Ch 14 #107
To relate pH to the equations we learned, E° has to be 0 when pH 7 since pH=7 when concentration of H+ and OH- are equal to 1.0 x 10^-7 mol.L^-1 when k=1 E°is 0 take ∆G°=-nFE°and ∆G°=RTlnK and then you get E°= RT/nF (lnK) at pH=7: K= [H+]/[OH-]= 1.0 x 10^-7 mol.L^-1/1.0 x 10^-7 mol.L^-1= 1 and E = 0...
- Mon Feb 27, 2017 9:49 pm
- Forum: General Science Questions
- Topic: Chemisty 14Bl and Chemistry 14C
- Replies: 2
- Views: 779
Re: Chemisty 14Bl and Chemistry 14C
There are a lot of students in my BL that are in 14C currently and many that haven't taken 14B in over a year and they're doing well!
- Tue Feb 14, 2017 12:35 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 4036231
Re: Post All Chemistry Jokes Here
Happy Valentine's Day!
- Sun Feb 12, 2017 9:00 pm
- Forum: Student Social/Study Group
- Topic: Chemistry Joke
- Replies: 57
- Views: 5487
- Mon Feb 06, 2017 5:08 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Homework 14.37
- Replies: 1
- Views: 500
Re: Homework 14.37
You would use E˚cell = E˚cathode - E˚anode with E˚ at the electrodes being the the electron potential for the half reaction. 2H+ (aq, 1.0 M) +2e- ----> H2 (g, 1 atm) E˚cathode = 0 V H2 (g, 1 atm) ----> 2H+ (aq, .075 M) +2e- E˚anode = 0 V 2H+ (aq, 1.0 M) +2e- + H2 (g, 1 atm) ----> 2H+ (aq, .075 M) +2...
- Sun Feb 05, 2017 4:07 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 4036231
- Sun Jan 29, 2017 3:19 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 4036231
Re: Post All Chemistry Jokes Here
alcohol is not a problem it's a solution
- Thu Dec 01, 2016 8:01 pm
- Forum: Ionic & Covalent Bonds
- Topic: Polarizability
- Replies: 2
- Views: 571
Re: Polarizability
1) if it is near an anion which has a high charge density (small and high charge)
2) if the cation is large since its electrons will be further away from the protons in its nucleus
2) if the cation is large since its electrons will be further away from the protons in its nucleus
- Thu Dec 01, 2016 7:56 pm
- Forum: Bronsted Acids & Bases
- Topic: Corrosive Properties
- Replies: 1
- Views: 609
Re: Corrosive Properties
Acids tend to accept electrons, and bases tend to donate electrons. Acids contain H+ ions. They tend to grab electrons and form hydrogen gas and since metals don't have a tight grip on their electrons. They give them up and the solid metal gradually disappears. The reaction is Bases contain OH- ions...
- Mon Nov 28, 2016 8:06 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: HW #12.69
- Replies: 1
- Views: 576
Re: HW #12.69
Al3+ forms a complex with water as ligands and since Al3+ is a Lewis acid AlCl3 gets substituted by Al(H2O)6
- Mon Nov 28, 2016 7:46 pm
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: Chapter 12 Problem 117
- Replies: 1
- Views: 498
Re: Chapter 12 Problem 117
A chem moderator said to omit 12.117 :)
- Mon Nov 28, 2016 7:34 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: When to use atm v. bar units [ENDORSED]
- Replies: 4
- Views: 2033
Re: When to use atm v. bar units [ENDORSED]
It would specify if the problem wanted the solution to be in bars
- Sat Nov 26, 2016 12:31 pm
- Forum: Amphoteric Compounds
- Topic: Basic, Acidic or Amphoteric [ENDORSED]
- Replies: 4
- Views: 836
Re: Basic, Acidic or Amphoteric [ENDORSED]
As2O3 is amphoteric because it can react with either acid or base.
Ex. Reaction with Acid
As2O3 we have: As2O3 + 6HCl --> 2AsCl3 + 3H2O
Reaction with base:
NaOH + As2O3 + 3H2O --> 2Na[As(OH)4]
Ex. Reaction with Acid
As2O3 we have: As2O3 + 6HCl --> 2AsCl3 + 3H2O
Reaction with base:
NaOH + As2O3 + 3H2O --> 2Na[As(OH)4]
- Sat Nov 26, 2016 12:55 am
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: 11.67
- Replies: 1
- Views: 581
Re: 11.67
Reactants are strongly favored so we push the reaction as far to the left as possible and then start from the new initial conditions Pressures (bar) 2 HCl(g) -> H2(g) + Cl2(g) Original 2.0 1.0 3.0 New initial 4.0 0.0 2.0 Change -2x +x +x Final 4.0-2x +x 2.0-x K= PH2 PCl2 / PHCl2 3.2 x 10 ^-34 = x(2....
- Mon Nov 14, 2016 11:12 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Equilibrium Concentration Units [ENDORSED]
- Replies: 3
- Views: 1656
Re: Equilibrium Concentration Units [ENDORSED]
You would use molarity as the units when calculating equilibrium concentrations!
- Mon Nov 14, 2016 11:07 am
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 4036231
- Tue Nov 01, 2016 11:45 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 4036231
Re: Post All Chemistry Jokes Here
Accurately portrays my love life
- Sun Oct 30, 2016 4:01 pm
- Forum: Lewis Structures
- Topic: H/W Q: 3.51- The Lewis stucture of HClO?
- Replies: 1
- Views: 861
Re: H/W Q: 3.51- The Lewis stucture of HClO?
HClO is an oxoacid and in oxoacids hydrogen has to be bonded with oxygen.
- Thu Oct 20, 2016 10:55 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 4036231
Re: Post All Chemistry Jokes Here
A photon checks into a hotel and is asked if he needs an help with his luggage.
He replies " No, I'm traveling light."
He replies " No, I'm traveling light."
- Sat Oct 15, 2016 3:48 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: 3d and 4s orbitals
- Replies: 2
- Views: 1115
Re: 3d and 4s orbitals
3D is only lower in energy than 4s after the 4s orbital is occupied
- Thu Oct 06, 2016 4:32 pm
- Forum: Properties of Light
- Topic: Atomic Spectra- Question 1.15 [ENDORSED]
- Replies: 3
- Views: 918
Re: Atomic Spectra- Question 1.15 [ENDORSED]
I solved this problem using the Rydberg equation instead of the bohr frequency condition.
You solved the n1=1 so for a line that's 102.6 nm you use the formula v=c/(lambda) = 2.922 x10^15 s^-1
Then use the Rydberg to solve for n2 which would give you n2=3
You solved the n1=1 so for a line that's 102.6 nm you use the formula v=c/(lambda) = 2.922 x10^15 s^-1
Then use the Rydberg to solve for n2 which would give you n2=3
- Sun Oct 02, 2016 10:23 pm
- Forum: Student Social/Study Group
- Topic: CHEM JOKES
- Replies: 29
- Views: 7061
Re: CHEM JOKES
Do you have 11 protons? Cause your sodium fine.