Search found 61 matches
- Sat Mar 16, 2019 10:24 am
- Forum: General Rate Laws
- Topic: Question 15.55 (Sixth Edition)
- Replies: 1
- Views: 225
Re: Question 15.55 (Sixth Edition)
a is true because the equilibrium constant K is equal to k/k' which is the forward rate constant divided by the reverse rate constant. So if K is very large, that implies that k is much larger than k', showing that the forward rate constant is much larger than the reverse rate constant. b is false b...
- Sat Mar 16, 2019 10:20 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 6th Edition 11.7
- Replies: 1
- Views: 288
Re: 6th Edition 11.7
For me, I noticed that the third and fourth flasks have the exact same number of diatomic molecules and single atoms, so that means that the reactants aren't decreasing anymore and the products are being made anymore at flask three. So that is the one where equilibrium was first reached.
- Sat Mar 16, 2019 10:11 am
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Anode and Cathode determination
- Replies: 1
- Views: 413
Re: Anode and Cathode determination
The cathode is the half-reaction with the most positive E while the anode is the half-reaction with the most negative E. This is so that the Ecell would be positive and that the reaction will occur
- Sat Mar 16, 2019 10:09 am
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Ecell
- Replies: 3
- Views: 1992
Re: Ecell
A more positive Ecell means that the reaction is more spontaneous. So if the Ecell was negative, the reaction would not occur, and the cell would not properly react.
- Sat Mar 16, 2019 10:07 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Cell Diagram - Water
- Replies: 2
- Views: 270
Re: Cell Diagram - Water
No, H2O(l) is not included in a cell diagram.
- Fri Mar 15, 2019 5:05 pm
- Forum: First Order Reactions
- Topic: 15.37
- Replies: 1
- Views: 435
Re: 15.37
For this question you would want to plug into the integrated rate law for the first order which is ln([A]t/[A]0) = -kt. In part c, [A]t would be 0.10[A]0 and so you can just plug that into the equation to find the time needed. The same thing applies for part d
- Fri Mar 15, 2019 4:56 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Lecture Cell Diagram Example
- Replies: 3
- Views: 267
Re: Lecture Cell Diagram Example
Basically, the two solids aren't completely mixed together because they are solid and can physically be separated, so they wouldn't need a comma to separate them. Compounds/molecules in aqueous/gas phases would have a comma and not a line because these phases are mixed and in the same "phase&qu...
- Fri Mar 15, 2019 4:54 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: H+ in Cell Diagram
- Replies: 3
- Views: 366
Re: H+ in Cell Diagram
Yeah you should always be putting H+ in the cell diagram if it is present in the total reaction. As far as I know, water is the only molecule that you don't include in the cell diagram.
- Fri Mar 15, 2019 4:53 pm
- Forum: First Order Reactions
- Topic: pseudo first order
- Replies: 2
- Views: 350
Re: pseudo first order
I'm pretty sure that the question will state that one is obviously "very large" or in excess so that a pseudo first order reaction can be conducted.
- Fri Mar 15, 2019 1:25 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Ways to practice pre-equilibrium approach
- Replies: 1
- Views: 234
Re: Ways to practice pre-equilibrium approach
If you have time and the resources, it would be nice to probably go over past tests in a course reader or from a test bank. It might also be good to check out the few review sessions this weekend to see if they have problems on pre-equilibrium approach
- Fri Mar 15, 2019 1:17 pm
- Forum: First Order Reactions
- Topic: Half-life Formulas
- Replies: 2
- Views: 548
Re: Half-life Formulas
I'm pretty sure you only need to know which formula goes with which order, but I mean, understanding how to get the half-life would probably help a lot not only on the test but on conceptual problems as well.
- Fri Mar 15, 2019 1:15 pm
- Forum: First Order Reactions
- Topic: Exponents for rate laws
- Replies: 1
- Views: 230
Re: Exponents for rate laws
The coefficient only determines the exponent in an elementary step. The exponent of a rate law IS the order, but you find it by identifying the slow elementary step in a reaction and going from there.
- Fri Mar 15, 2019 1:14 pm
- Forum: Balancing Redox Reactions
- Topic: Balanced Half Reactions
- Replies: 5
- Views: 666
Re: Balanced Half Reactions
I'm pretty sure balanced half reactions would be only considering the atoms and charge of that half reaction, so it would include balancing oxygens and hydrogens by using water and H+ or OH-. But I think balancing the charge between a reduction and oxidation reactions is for the overall reaction, an...
- Fri Mar 15, 2019 10:11 am
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Largest E Cell
- Replies: 2
- Views: 685
Re: Largest E Cell
You find the largest E Cell by subtracting the E of the molecule with the highest oxidizing power (high reduction potential; most positive; in the cathode) by the E of the molecule with the highest reducing power (high oxidation potential; most negative; in the anode).
- Fri Mar 15, 2019 10:08 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Adding Pt(s) to cell diagram
- Replies: 5
- Views: 509
Re: Adding Pt(s) to cell diagram
Add Pt(s) to a cell diagram on the side that does not have a conducting solid. So each side must have a solid that is a conductor (usually look for metals).
- Fri Mar 15, 2019 10:07 am
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Exothermic Graph
- Replies: 2
- Views: 267
Re: Exothermic Graph
This graph is showing a two-step reaction in which the second step is the slower step. So I'm not exactly sure what ES stands for, but that should be where the intermediates are made for the second step. So the first hump is the activation energy needed for the first step and the second hump is the ...
- Fri Mar 15, 2019 10:04 am
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: 6th edition 15.19 part c
- Replies: 1
- Views: 214
Re: 6th edition 15.19 part c
It probably isn't necessary, but I'm pretty sure that on the test it would be specific as to what unit is needed.
- Fri Mar 15, 2019 10:03 am
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: 6th edition 15.19 part c
- Replies: 1
- Views: 403
Re: 6th edition 15.19 part c
It doesn't matter which reaction you use to solve for k; they all have the same rate constant.
- Fri Mar 15, 2019 10:01 am
- Forum: Second Order Reactions
- Topic: 6th edition 15.39 part b
- Replies: 2
- Views: 476
Re: 6th edition 15.39 part b
This is because the product concentrations are implied to be at 0 when the reaction starts. So logically, as product is made, the reactant is used up and decreases. So in this case, because B increases to 0.19, A would decrease by 0.095 (if you look at the stoichiometric coefficients). So then, the ...
- Fri Mar 15, 2019 9:58 am
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: 6th edition 15.17
- Replies: 1
- Views: 395
Re: 6th edition 15.17
When C is independent of the rate, that means that it has zero order. It's not part of the rate law, but [C] is still considered zero order. So when you do experimental analysis, you do ignore [C] when calculating the orders of [A] and [B].
- Fri Mar 15, 2019 9:55 am
- Forum: Second Order Reactions
- Topic: 3rd order and fractional orders
- Replies: 1
- Views: 481
Re: 3rd order and fractional orders
Because the order is based on experimental analysis, you can use the same method as you would if the order was an integer. So, you would find the factor of change in both the initial rate and the reactant concentration. and then you would divide the natural log of the change in initial rate by the n...
- Fri Mar 15, 2019 9:47 am
- Forum: Second Order Reactions
- Topic: Problem 51 6th Edition
- Replies: 1
- Views: 409
Re: Problem 51 6th Edition
The order is actually two because the overall rate law is based on the slow step, the slow step being NO + Br2 --> NOBr2. So, the rate law would be k[NO][Br}, which is a second order reaction.
- Fri Mar 15, 2019 9:45 am
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Rate constant for large equilibrium constant
- Replies: 1
- Views: 233
Re: Rate constant for large equilibrium constant
Yes, because the equilibrium constant equals the rate of the forward reaction over the rate of the reverse reaction, so if the equilibrium constant is very large, then that implies that the rate of the forward reaction is much greater than the rate of the reverse reaction.
- Fri Mar 15, 2019 9:43 am
- Forum: General Rate Laws
- Topic: Determining the order
- Replies: 3
- Views: 501
Re: Determining the order
No, only if the reaction is said to be an elementary step can you use the coefficients to accurately depict the order of the step. But in the case of a general reaction, no, you would need more information besides the balanced equation to find out the order of the reaction
- Fri Mar 15, 2019 9:42 am
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: rate constant
- Replies: 1
- Views: 228
Re: rate constant
This statement should be false because the rate constant doesn't change based on concentration of reactants. The rate in total does change, but the rate constant changes based on the Arrhenius equation, which involves temperature, activation energy, and collision rate, but NOT concentration.
- Fri Mar 15, 2019 9:40 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: ∆H and q [ENDORSED]
- Replies: 6
- Views: 709
Re: ∆H and q [ENDORSED]
It equals 0 under constant pressure.
- Thu Mar 14, 2019 9:40 pm
- Forum: First Order Reactions
- Topic: Edition 6 15.23
- Replies: 2
- Views: 304
Re: Edition 6 15.23
Actually, I believe that the coefficients don't necessarily determine the order of the entire reaction because it ultimately boils down to the experimental analysis of initial rates to find the order (unless it is already given). The coefficient rule applies to elementary steps but not if the questi...
- Mon Feb 25, 2019 10:24 pm
- Forum: Balancing Redox Reactions
- Topic: Question 14.5a 6th Edition [ENDORSED]
- Replies: 1
- Views: 197
Question 14.5a 6th Edition [ENDORSED]
The first chemical reaction is O3 + Br- --> O2 + BrO3- and it asks to balance this reaction. My question is that the solution manual states that the reduction half-reaction is O3 --> O2 but when I was assigning oxidation numbers to all the elements, the oxygen atom in BrO3- had a lower oxidation num...
- Mon Feb 25, 2019 6:59 pm
- Forum: Balancing Redox Reactions
- Topic: Balancing Electrons
- Replies: 3
- Views: 316
Re: Balancing Electrons
You look at the charge on both sides and increase the amount of electrons to make the charges on both sides equal.
- Mon Feb 25, 2019 6:57 pm
- Forum: Balancing Redox Reactions
- Topic: Balancing Redox Reactions
- Replies: 3
- Views: 362
Re: Balancing Redox Reactions
The side with excess H2O molecules should have the leftover H2O molecules, but overall it shouldn't matter which side the molecules are on.
- Sat Dec 08, 2018 12:17 am
- Forum: Naming
- Topic: Bis, Tris, Tetrakis, etc.
- Replies: 3
- Views: 1103
Re: Bis, Tris, Tetrakis, etc.
Bis-, Tris-, and Tetrakis- are typically used for larger ligand to show that there is 2/3/4 of the entire ligand and not just one part of it. A good example would be using these prefixes with ethlyenediamine which is a large ligand.
- Sat Dec 08, 2018 12:11 am
- Forum: Naming
- Topic: Oxidation Number
- Replies: 2
- Views: 265
Re: Oxidation Number
You find the oxidation number of the metal cation by looking at the ligands and seeing whether they have a charge or not. Then you see what the overall charge of the entire compound is to predict what the leftover charge of the cation would be. The oxidation number of the metal cation + the charge o...
- Sat Dec 08, 2018 12:09 am
- Forum: Sigma & Pi Bonds
- Topic: Why are sigma and pi bonds drawn differently?
- Replies: 1
- Views: 486
Re: Why are sigma and pi bonds drawn differently?
Sigma bonds are specifically for single bonds, and they overlap as they do to show that they can rotate without the bonds breaking. Pi bonds are for the second and third bonds on double and triple bonds respectively. These bonds are parallel to each other and overlap side to side to show that the bo...
- Sat Dec 08, 2018 12:06 am
- Forum: Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)
- Topic: Induced dipole?
- Replies: 3
- Views: 488
Re: Induced dipole?
An induced dipole is when an atom with a regularly neutral charge is affected by a nearby atom/ion with a high charge that causes it to become slightly positive/negative. It affects London Dispersion forces because LDFs are pretty much when two induced dipoles are attracted to each other.
- Sat Dec 08, 2018 12:04 am
- Forum: Naming
- Topic: Naming (churro)
- Replies: 3
- Views: 329
Re: Naming (churro)
Chlorine is labeled as chloro because as a ligand, you change the name and add an o to ligands that are single atoms/usually ions (ex. bromo, iodo). The bis is to imply that there is two of the entire ethylenediamine ligand. Usually with smaller ligands, we just use di, tri, tetra etc, but with bigg...
- Fri Dec 07, 2018 11:58 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Ligand
- Replies: 1
- Views: 223
Re: Ligand
I'm assuming they won't all be on the test and they won't ask us to specifically identify ligands, but I think it would be very, very important and useful to memorize them for coordination compound questions.
- Fri Dec 07, 2018 11:56 pm
- Forum: Hybridization
- Topic: Expanded Octet Hybridizations
- Replies: 2
- Views: 447
Re: Expanded Octet Hybridizations
In general, we use the d orbitals for expanded octets because using s and p orbitals, you can only reach a maximum of 8 electrons, which is an octet. So we use d orbitals for when molecules have an expanded octet. In hybridization, we use d orbitals when s and p orbitals aren't enough for all the va...
- Fri Dec 07, 2018 11:50 pm
- Forum: Ionic & Covalent Bonds
- Topic: Boiling point
- Replies: 3
- Views: 829
Re: Boiling point
C5H12 has a higher boiling point because as the molecule increases in size, there is more surface area for Van Der Waals forces to attract the molecule, and thus making the IMF of C5H12 molecule stronger and the boiling point higher.
- Fri Dec 07, 2018 11:15 pm
- Forum: Coordinate Covalent Bonds
- Topic: polydentates
- Replies: 5
- Views: 583
Re: polydentates
I feel like there's not really an "easy" way to find out whether a compound is a polydentate, but I think the best way is to write out the Lewis diagram and see whether there are multiple locations where the ligand can bond with the metal cation. Also take into account the sigma and pi bon...
- Tue Nov 27, 2018 12:34 pm
- Forum: Lewis Structures
- Topic: Electron Configuration of N5+ Cation [ENDORSED]
- Replies: 1
- Views: 962
Electron Configuration of N5+ Cation [ENDORSED]
In the 6th edition textbook, question 99 asks to find the electron configuration for a salt N5+, and I found that using single and double bonds and several lone pairs would make most of the atoms in the molecule have a formal charge of 0 while each atom fulfills the octet rule. So how come the answe...
- Mon Nov 19, 2018 5:25 pm
- Forum: Lewis Structures
- Topic: Lewis Structure of Carbon Monoxide
- Replies: 3
- Views: 433
Lewis Structure of Carbon Monoxide
Is there a reason carbon monoxide has a triple bond between the carbon and oxygen atoms? Aren't the formal charges more attractive when there's a double bond between them with two lone pairs on oxygen and one lone pair on carbon?
- Mon Nov 19, 2018 5:22 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Molecular geometry vs Electron geometry
- Replies: 6
- Views: 701
Re: Molecular geometry vs Electron geometry
Yeah! The electron geometry is pretty much how many electron densities are around the central atom, regardless of whether the electron densities are from a bond or from a lone pair. So, for instance, ammonia (NH3) has an electron geometry of a tetrahedral because it has 3 bonds with the hydrogen ato...
- Sun Nov 18, 2018 3:56 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Electron Configuration of Rutherfordium
- Replies: 2
- Views: 383
Electron Configuration of Rutherfordium
I understand that we will not have to cover most of the lower transition metals on our tests, but on homework question 45 of section 2 in the 6th edition, the electron configuration [Rn]7s^2 6d^2 gives Thorium. How come? Shouldn't Rutherfordium fit best with that electron configuration? Isn't Thoriu...
- Sun Nov 18, 2018 3:51 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Molecular geometry vs Electron geometry
- Replies: 6
- Views: 701
Re: Molecular geometry vs Electron geometry
Yes, as we covered both topics pretty in depth in lecture several times.
- Sun Nov 18, 2018 3:50 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: atom orientation
- Replies: 1
- Views: 109
Re: atom orientation
It depends on which orientation has the most distance between neighboring atoms as possible, if that makes sense. So, trigonal planar is in one plane because the neighboring atoms are farthest apart when in one plane. However, a tetrahedral is in multiple planes because again, the neighboring atoms ...
- Sun Nov 18, 2018 3:43 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Nickel Electron Configuration
- Replies: 3
- Views: 4887
Nickel Electron Configuration
On number 51 of Chapter 2, 6th edition, the amount of unpaired electrons of nickel was said to be 2 unpaired electrons, but I'm not exactly sure why. When looking at the periodic table, shouldn't the amount of unpaired electrons be 3?
- Sun Nov 11, 2018 3:27 pm
- Forum: Properties of Light
- Topic: Wave-Particle Duality
- Replies: 1
- Views: 289
Wave-Particle Duality
Which experiments display the wave characteristics of a photon and which ones display the particle characteristics?
- Sun Nov 11, 2018 3:22 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Lone Pairs
- Replies: 7
- Views: 699
Lone Pairs
Do the lone pairs of the atoms not at the center of the molecule affect the shape at all? Don't the lone pairs also contribute to electron
repulsion and thus shape as well?
repulsion and thus shape as well?
- Sun Nov 11, 2018 3:14 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: molecular shapes
- Replies: 13
- Views: 906
Re: molecular shapes
I'm positive that we will go over a lot of other shapes because we haven't gone over shapes like bent shape, which is the shape of H2O.
- Thu Nov 01, 2018 3:01 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Diagonal Relationships
- Replies: 1
- Views: 165
Diagonal Relationships
Diagonal relationships weren't covered in Professor Lavelle's lectures, but there was homework assigned in that topic,
so would that be on the test? If so, do we need to know specific examples of diagonal relationships or how they arise?
so would that be on the test? If so, do we need to know specific examples of diagonal relationships or how they arise?
- Thu Nov 01, 2018 2:44 pm
- Forum: Trends in The Periodic Table
- Topic: Electron Affinity
- Replies: 5
- Views: 556
Electron Affinity
I understand the concept of electron affinity being the energy released when an atom receives an electron and that the trend is that electron affinity generally increases across a period and decreases down a group. However, I was wondering why more energy is released from atoms in the top right of t...
- Tue Oct 30, 2018 9:02 pm
- Forum: Lewis Structures
- Topic: Central Atom
- Replies: 13
- Views: 818
Re: Central Atom
For any molecules, the central atom is the atom with the lowest ionization energy, so in the case of CH3OH, C would have the lowest
ionization energy, thus being in the center of the molecule.
ionization energy, thus being in the center of the molecule.
- Sun Oct 28, 2018 4:28 pm
- Forum: Ionic & Covalent Bonds
- Topic: the Octet rule
- Replies: 21
- Views: 3904
Re: the Octet rule
As far as I know, most atoms should follow the octet rule. There are a few atoms (mostly unstable ones) that bond in
very odd formations, but I'm sure we will go over them in class or they won't be important for the course.
very odd formations, but I'm sure we will go over them in class or they won't be important for the course.
- Sun Oct 28, 2018 4:02 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: 3d and 4d
- Replies: 1
- Views: 272
Re: 3d and 4d
This is based mainly on the principal quantum number because the main energy differences between different electrons
are based on the quantum numbers. So for example, all electrons in quantum level 3 would have less energy than
electrons in quantum level 4.
are based on the quantum numbers. So for example, all electrons in quantum level 3 would have less energy than
electrons in quantum level 4.
- Tue Oct 23, 2018 2:10 am
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Bohr Frequency Condition
- Replies: 2
- Views: 286
Bohr Frequency Condition
I understand that in the equation E = - h R / n^2, E is symbolizing the change in energy, but would using the E from E = hv be equal to that change in energy? and what does the n stand for in that equation? The initial or final energy level?
- Sun Oct 21, 2018 5:25 pm
- Forum: Properties of Electrons
- Topic: Wave Model
- Replies: 1
- Views: 217
Re: Wave Model
It's kind of hard to explain, but I'll try my best. The quantized energy levels are shown in the wave model because only with specific allowed energies can there be an accurate wave model where the ends of the waves eventually meet. If there's an energy that isn't "allowed", then the ends ...
- Sun Oct 21, 2018 3:13 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Diffraction and Spectral Lines
- Replies: 1
- Views: 122
Diffraction and Spectral Lines
Does diffraction, specifically constructive and destructive interference, have anything to do with the specific values of spectral lines? Or any correlation at all? More fundamentally, how are spectral lines calculated/found? I'm still a bit confused about the topic.
- Sun Oct 21, 2018 2:55 pm
- Forum: *Shrodinger Equation
- Topic: Application of Schrodinger's Equation
- Replies: 2
- Views: 346
Application of Schrodinger's Equation
What exactly do we use Schrodinger's Equation for? I understand we don't need the actual equation on the test, but what is its purpose when learning about wave properties? I didn't really understand the concept
- Mon Oct 01, 2018 2:22 am
- Forum: Molarity, Solutions, Dilutions
- Topic: Question G.13 Answer Error (Sixth Edition) [ENDORSED]
- Replies: 1
- Views: 186
Question G.13 Answer Error (Sixth Edition) [ENDORSED]
In the sixth edition of the textbook, the answer of G.13 is shown to be 1.0 x 10^-2 mol C when the question is clearly addressing nitrogen atoms in fertilizer. Just for the sake of checking answers, did anyone else receive 0.6 mol N as the answer?
- Mon Oct 01, 2018 12:41 am
- Forum: SI Units, Unit Conversions
- Topic: Formula Units: Molecules or Atoms
- Replies: 5
- Views: 435
Re: Formula Units: Molecules or Atoms
It mainly depends on whether the question specifically asks for molecules or atoms. Obviously, if the question refers to a single element, then it would be atoms. Otherwise, it mainly depends on what the question asks for.
- Sat Sep 29, 2018 12:38 am
- Forum: SI Units, Unit Conversions
- Topic: Addressing Hydrates in Conversion Problems
- Replies: 1
- Views: 146
Addressing Hydrates in Conversion Problems
In question E9(a) of the sixth edition textbook, it asks for the amount of oxygen atoms in magnesium sulfate pentahydrate. Does this include the oxygen atoms in the pentahydrate part of the solution?