Search found 29 matches

by Laura Rabichow 1J
Tue Mar 14, 2017 9:46 pm
Forum: Balancing Redox Reactions
Topic: Winter 2016 Midterm, Question 6
Replies: 2
Views: 383

Re: Winter 2016 Midterm, Question 6

There are two C molecules in the compound (C2H5OH), each going from 2- to 1-. It's a change of 1 electron per C and with 2 C in the compound that makes it a total change of 2 electrons.
by Laura Rabichow 1J
Mon Mar 13, 2017 10:01 pm
Forum: Balancing Redox Reactions
Topic: Help on practice test
Replies: 4
Views: 382

Re: Help on practice test

Like Chem_Mod said its 2 electrons lost for Fe between reactants and products. The reaction uses 2 moles of Fe, so its 2 moles Fe times 2 moles electrons per Fe, thus making n = 4.
by Laura Rabichow 1J
Mon Mar 13, 2017 9:52 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: q=mc∆t vs q=c∆t? (chapter 8)
Replies: 2
Views: 9537

Re: q=mc∆t vs q=c∆t? (chapter 8)

In general, you can figure it out based on the units of the given C. You typically want q to be in kJ or J. Like Jessica said, in your example problem the C was given in kJ/˚C, so you would only multiply by the temperature. If your C was given in kJ/(˚C*mol), you would use q = nC∆T. If your C was gi...
by Laura Rabichow 1J
Tue Mar 07, 2017 10:45 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: Transition State and Psuedo-Equilibrium Rate Constant
Replies: 1
Views: 268

Re: Transition State and Psuedo-Equilibrium Rate Constant

Regarding page 85, the we know that a larger ∆G means that the reaction will have a higher activation energy. Step 1 shows the transition state TS1 and step 2 shows the transition state TS2. On the graph, you'll see that TS1 has a higher free energy peak than TS2 and thus has a higher energy barrier...
by Laura Rabichow 1J
Tue Mar 07, 2017 10:30 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Why is K = kfor/krev?
Replies: 3
Views: 574

Re: Why is K = kfor/krev?

We were told that the rate of the forward reaction is equal to the rate of the reverse reaction in a given chemical equation. For example, if you have A + B -> C + D, then kfor[A][B] = krev[C][D]. If you divide the right side by [A][B] and the left side by krev, you get kfor/krev = [C][D]/[A][B], or...
by Laura Rabichow 1J
Mon Feb 27, 2017 8:48 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Cell potential
Replies: 4
Views: 553

Re: Cell potential

Figure out which one is the anode and which is the cathode, then do Ecathode - Eanode to get the standard cell potential.
by Laura Rabichow 1J
Sat Feb 25, 2017 11:41 am
Forum: *Enzyme Kinetics
Topic: Catalyst [ENDORSED]
Replies: 2
Views: 325

Re: Catalyst [ENDORSED]

Yes, because they effect how quickly a reaction occurs.
by Laura Rabichow 1J
Fri Feb 24, 2017 1:37 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: Heterogenous catalysts
Replies: 1
Views: 257

Re: Heterogenous catalysts

Catalysts only lower the activation energy required to start the reaction, it doesn't effect whether or not a reaction is exothermic. Also, the number of molecules a catalyst binds to is dependent on the catalyst, not whether its heterogenous or homogenous. Heterogenous just means that its in a diff...
by Laura Rabichow 1J
Mon Feb 20, 2017 10:33 pm
Forum: General Rate Laws
Topic: Rates
Replies: 1
Views: 300

Re: Rates

Average rate is the change in concentration over some chosen interval of time, like say delta(t) = 5 sec (delta[P]/delta(t)). Instantaneous rate is the rate of change at an exact moment, which is equal to the slope at that point (d[P]/dt).
by Laura Rabichow 1J
Wed Feb 15, 2017 12:10 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Book problem 14.55
Replies: 2
Views: 297

Re: Book problem 14.55

It depends on the half-reaction, which I'm not able to look up at the moment. It's likely that you have to flip the half reaction, making E = -1.23 V. Then Etot = E(cathode) - E(anode), and a positive Etot is preferred.
by Laura Rabichow 1J
Fri Feb 10, 2017 3:39 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: Midterm Winter 2013 problem 5B
Replies: 10
Views: 917

Re: Midterm Winter 2013 problem 5B

CF4 is symmetrical, whereas the other three are asymmetrical. Asymmetry gives the other elements different possible structures, which gives them higher standard molar entropy. Furthermore, CH3F has the greatest # of possible arrangements, followed by CH2F2, then CHF3, which is why they're ranked in ...
by Laura Rabichow 1J
Fri Feb 03, 2017 2:01 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: Gibbs Free Energy and Electrochemistry
Replies: 2
Views: 347

Re: Gibbs Free Energy and Electrochemistry

Page 49 of the course reader may explain this better, but basically the maximum work (wmax) is related to the cell potential E by E = -w/charge. Rearranging gives you -charge x E = w. Faraday's constant gives the charge of 1 mol of electrons and multiplying by the number of moles of electrons transf...
by Laura Rabichow 1J
Thu Feb 02, 2017 7:03 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Units for Bond Enthalpy Questions
Replies: 2
Views: 327

Re: Units for Bond Enthalpy Questions

The course reader does use kJ as opposed to kJ/mol when multiplying by the number of moles.
by Laura Rabichow 1J
Sat Jan 28, 2017 9:32 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Molar Kinetic Energy
Replies: 1
Views: 252

Re: Molar Kinetic Energy

Probably 8.314 J/(K*mol) for that specific equation, just make sure your T values are in kelvin and you'll get Utot in J/mol.
by Laura Rabichow 1J
Fri Jan 20, 2017 3:19 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Kekulé structure?
Replies: 1
Views: 225

Re: Kekulé structure?

A kekule structure is just when the bonds in a compound are drawn as lines, like a benzene ring. I don't have the book in front of me so it should just be referring to how the compound in the problem is drawn, but I'm attaching a photo of benzene as an example. Screen Shot 2017-01-20 at 3.18.24 PM.p...
by Laura Rabichow 1J
Sat Jan 14, 2017 3:36 pm
Forum: Phase Changes & Related Calculations
Topic: Negative work values
Replies: 1
Views: 270

Re: Negative work values

A negative value just indicates that the force is being applied in a direction opposite to movement. So, for example, let's say I have a container filled with gas and a piston at the top. If the gas starts to expand (like if I increased temperature or something) and the piston moves upward, but I pu...
by Laura Rabichow 1J
Tue Jul 26, 2016 11:44 pm
Forum: Calculating the pH of Salt Solutions
Topic: Spectator Ions?
Replies: 5
Views: 886

Re: Spectator Ions?

Na+ is a group 1 cation, so yes.
by Laura Rabichow 1J
Tue Jul 26, 2016 11:41 pm
Forum: Conjugate Acids & Bases
Topic: Conjugate Acids & Conjugate Bases? [ENDORSED]
Replies: 2
Views: 922

Re: Conjugate Acids & Conjugate Bases? [ENDORSED]

If you have an acid such that HA <-> H+ + A-, then A- is the conjugate base of HA. If you have a base such that BOH <-> B+ + OH-, then B+ is the conjugate acid. Remember that by bronsted definitions, an acid is a proton donor and a base is a proton acceptor. Using my HA example, since I put arrows t...
by Laura Rabichow 1J
Tue Jul 26, 2016 11:22 pm
Forum: *Indicators
Topic: Indicators
Replies: 3
Views: 946

Re: Indicators

Polyprotic acids and bases are also excluded
by Laura Rabichow 1J
Mon Jul 25, 2016 8:22 pm
Forum: Conjugate Acids & Bases
Topic: Equilibrium sign or not [ENDORSED]
Replies: 4
Views: 590

Re: Equilibrium sign or not [ENDORSED]

The same applies for a strong base.
by Laura Rabichow 1J
Mon Jul 25, 2016 8:20 pm
Forum: Properties & Structures of Inorganic & Organic Acids
Topic: Lewis structures of acids and bases [ENDORSED]
Replies: 2
Views: 455

Re: Lewis structures of acids and bases [ENDORSED]

It's not based on the lewis dot structure. It's based on whether the solution becomes acidic or basic, so for example HCl gives off an H+ in an aqueous solution, forming an H30+ that makes the solution more acidic. You either know based on the atoms in the molecule (i.e. HCl can't give off an OH- bu...
by Laura Rabichow 1J
Sat Jul 23, 2016 10:41 pm
Forum: Lewis Acids & Bases
Topic: Lewis Acid and Base [ENDORSED]
Replies: 2
Views: 295

Re: Lewis Acid and Base [ENDORSED]

They both are capable of receiving an electron pair. For example, if you were to throw NaOH into a solution that already had H+, the H+ can accept the electron pair from the OH-
by Laura Rabichow 1J
Sat Jul 23, 2016 10:35 pm
Forum: Naming
Topic: Mono-, Bi-, Polydentate Ligands [ENDORSED]
Replies: 1
Views: 761

Re: Mono-, Bi-, Polydentate Ligands [ENDORSED]

Based on some previous exam questions, I think we do need to know which ligands are mono, bi, and polydentate. For example, from the Fall 2010 final exam in the course reader question 4B asks for the shape of [Co(en)2(Br)2], which you only know if you understand that (en) is polydentate.
by Laura Rabichow 1J
Wed Jul 20, 2016 11:38 pm
Forum: Naming
Topic: Naming complex compounds
Replies: 5
Views: 1443

Re: Naming complex compounds

You do need to memorize that chart in the course reader. In it, neutral ligands (ethylenediamine, ammine, etc) are listed and have oxidation numbers of 0. For the charged ligands the oxidation numbers are listed in the table as well.
by Laura Rabichow 1J
Mon Jul 18, 2016 12:44 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: chemical Equilibrium [ENDORSED]
Replies: 2
Views: 586

Re: chemical Equilibrium [ENDORSED]

Calculate Q by using the concentrations given. So Q = [CO2]/[CO]^2. If Q=K, the reaction is in equilibrium. If Q<K, then there are more reactants than products and so the reaction proceeds right. If Q>K, there are more products than reactants and so the reaction proceeds left.
by Laura Rabichow 1J
Fri Jul 15, 2016 6:47 pm
Forum: Hybridization
Topic: HW 4.93: Hybridization of Carbon Atoms in Cyclopropane (C6H6 [ENDORSED]
Replies: 3
Views: 734

Re: HW 4.93: Hybridization of Carbon Atoms in Cyclopropane ( [ENDORSED]

IMG_5484.jpg The carbon atoms are in a ring, not a straight line. Also, the molecular formula is C3H6 instead of C6H6. I think that may be where the mixup is occurring. Since each carbon is attached to each other, that uses up two out of four bonds for each of them. Each carbon then has two hydroge...
by Laura Rabichow 1J
Thu Jul 07, 2016 11:35 am
Forum: Formal Charge and Oxidation Numbers
Topic: Q5b, 2012 Midterm - Formal Charge and Greenhouse Gases [ENDORSED]
Replies: 1
Views: 326

Re: Q5b, 2012 Midterm - Formal Charge and Greenhouse Gases [ENDORSED]

It's not the formal charges that make gases "greenhouse gases." Greenhouse gases that deplete the ozone layers are radicals, compounds that have a single unpaired electron in them. I don't have the practice exam in front of me right now so I don't know exactly which compound the question w...
by Laura Rabichow 1J
Fri Jul 01, 2016 2:44 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: The equals(=) sign or greater than (>) sign? [ENDORSED]
Replies: 3
Views: 497

Re: The equals(=) sign or greater than (>) sign? [ENDORSED]

Actually, there is one situation where you do use an = instead of > . If you are asked to find the minimum uncertainty (of either x, v, or p), then you simply use the equal sign since that gives you the minimum possible value. If you're not asked for the minimum value, but rather just the uncertaint...
by Laura Rabichow 1J
Sat Jun 25, 2016 3:11 pm
Forum: Properties of Light
Topic: Chapter 1 Quantum world HW
Replies: 1
Views: 259

Re: Chapter 1 Quantum world HW

I may be wrong, but here is how I interpreted the question: Lithium is a metal with a given work function of 2.93 eV. I'm assuming the way that the photomultiplier works is by detecting electrons emitted from the metal due to the beam of light consistently on it, and when that light is broken, no el...

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