Search found 60 matches
- Sun Mar 17, 2019 12:38 am
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: K vs k1/k'1
- Replies: 6
- Views: 1502
K vs k1/k'1
When writing the rate law using pre-equilibrium, does it matter if we write the answer as k2K or k2k1/k'1?
- Sun Mar 17, 2019 12:34 am
- Forum: Balancing Redox Reactions
- Topic: calculating standard cell potential
- Replies: 5
- Views: 808
Re: calculating standard cell potential
If you subtract cathode - anode, you don't flip the signs. If you add them, then you would have to flip the sign of the anode.
- Sun Mar 17, 2019 12:31 am
- Forum: Balancing Redox Reactions
- Topic: balaning redox reactions
- Replies: 3
- Views: 610
Re: balaning redox reactions
Yes, it is relevant when balancing the charges of the equation.
- Sat Mar 09, 2019 5:47 pm
- Forum: Zero Order Reactions
- Topic: Textbook Examples
- Replies: 1
- Views: 254
Re: Textbook Examples
In 6th edition 15.17 it involves a zero order reactant.
- Sat Mar 09, 2019 5:39 pm
- Forum: Zero Order Reactions
- Topic: Zero order
- Replies: 10
- Views: 1689
Re: Zero order
A zero order reaction is one where the rate is independent of the concentration. An example could be the decomposition of nitrous oxide on a hot platinum wire. The rate of that reaction doesn't depend on the concentration of nitrous oxide, it depends on whether surface area runs out on the wire for ...
- Sat Mar 09, 2019 5:28 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: 15.101 6th Edition
- Replies: 1
- Views: 271
Re: 15.101 6th Edition
I think it is because H20 is a liquid, so you don't include the concentration of it.
- Sat Mar 02, 2019 2:13 am
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Cell Potential Spontaneity
- Replies: 7
- Views: 870
Re: Cell Potential Spontaneity
You have to use the equation ∆G = -nFE. Looking at that relationship, you can see that when E is positive, ∆G is negative, and when E is negative, ∆G is positive. A reaction is spontaneous when ∆G is negative, and thus when E is positive.
- Sat Mar 02, 2019 2:11 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Order of Cell Diagrams
- Replies: 13
- Views: 1605
Order of Cell Diagrams
I know that in writing a cell diagram, the anode side goes first and then the cathode side goes next. But is there a way to know the order to write the terms of the anode and cathode? Like solids first then aqueous? Or does it not matter?
- Sat Mar 02, 2019 2:06 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Cell Diagrams (Using Platinum)
- Replies: 10
- Views: 1003
Re: Cell Diagrams (Using Platinum)
Platinum is used when there isn't a conducting solid. So in the textbook, there can be platinum on a side with a solid because that solid isn't a conductor. Platinum is used because it is inert and won't react.
- Sun Feb 24, 2019 4:14 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Electrodes
- Replies: 2
- Views: 368
Re: Electrodes
An inert conductor is used if there is no metal electrode. In this case copper is metal and an electrode, so an inert conductor would not be needed.
- Sun Feb 24, 2019 4:12 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Inert Conductors
- Replies: 3
- Views: 452
Re: Inert Conductors
Inert conductors transfer electrons, so in redox reactions they are the medium for the oxidation and reduction half-reactions.
- Sun Feb 24, 2019 4:11 pm
- Forum: Balancing Redox Reactions
- Topic: salt bridge?
- Replies: 3
- Views: 370
Re: salt bridge?
Since electrons are moving from the anode to the cathode, the salt bridge balances out the charges by donating negatively charged anions to the positively charged anode, and positively charged cations to the negatively charged cathode.
- Sun Feb 17, 2019 2:06 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Converting to q rev
- Replies: 4
- Views: 677
Re: Converting to q rev
Because entropy is a state function, the change will be the same regardless of whether the reaction is reversible or not, since it is not path dependent, so you can use q rev /T, as long as the reaction is isothermal. If it is isobaric (constant pressure), isochoric (constant volume), etc you would ...
- Sun Feb 17, 2019 1:36 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Gibbs Free Energy Signs
- Replies: 4
- Views: 472
Re: Gibbs Free Energy Signs
Negative ∆G means the forward reaction is spontaneous. Positive ∆G means the forward reaction is not spontaneous, but the reverse reaction is. If ∆G equals 0, that means the system is in equilibrium.
- Sun Feb 17, 2019 1:34 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Free Energy
- Replies: 1
- Views: 219
Re: Free Energy
Gibbs free energy is given by the equation ∆G = ∆H - T∆S An exothermic reaction is much more likely to be spontaneous than an endothermic reaction, as it releases heat instead of requiring it. The ∆H of an exothermic and likely spontaneous reaction is negative. A reaction that also increases entropy...
- Sun Feb 10, 2019 2:24 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.73 Help (Bond Enthalpy)
- Replies: 2
- Views: 346
Re: 8.73 Help (Bond Enthalpy)
For bond enthalpies, the C-C and C=C bonds of benzene are counted the same. So since benzene has 3 single carbon bonds and 3 double carbon bonds, it is a total of 6 C-C/C=C bonds formed.
- Sun Feb 10, 2019 2:21 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: hmwrk 8.21 6th edition
- Replies: 1
- Views: 261
Re: hmwrk 8.21 6th edition
The copper is losing heat to the water, so the equations would be set equal to each other. However, you would have to flip the sign of the copper because the heat lost would be negative, so you get -q copper = q water By using (100-T) you get rid of the negative sign so both values become positive. ...
- Sun Feb 10, 2019 2:16 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: Degeneracy
- Replies: 4
- Views: 542
Re: Degeneracy
You can determine degeneracy by looking at all the possible arrangements of the molecules, like if the positions of atoms changed.
- Mon Feb 04, 2019 2:26 am
- Forum: Calculating Work of Expansion
- Topic: Related to Question 5 of Ch.8
- Replies: 1
- Views: 227
Re: Related to Question 5 of Ch.8
The first equation is used when the reaction is an irreversible expansion against constant pressure. The second equation is used when the reaction is a reversible expansion.
- Mon Feb 04, 2019 2:24 am
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Work
- Replies: 6
- Views: 664
Re: Work
Work can be negative if it is done by the system, opposite of doing work on the system in that heat is released in contrast to adding heat, or energy.
- Mon Feb 04, 2019 2:21 am
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Isothermic definition
- Replies: 5
- Views: 909
Re: Isothermic definition
Isothermal is when temperature is constant. I think isometric is related to isomers, where molecules that can be superimposed after a reflection like in a mirror are called isomers.
- Sun Jan 27, 2019 2:27 pm
- Forum: Phase Changes & Related Calculations
- Topic: Delta H
- Replies: 11
- Views: 2016
Re: Delta H
Endothermic is positive ∆H because the enthalpy of the products is greater than the enthalpy of the reactants, so energy is absorbed. The opposite is true for exothermic, since energy is released.
- Sun Jan 27, 2019 2:15 pm
- Forum: Phase Changes & Related Calculations
- Topic: State Properties
- Replies: 3
- Views: 412
Re: State Properties
Heat is dependent on work due to the first law of thermodynamics, where ∆U = q + w, so q = ∆U - w. Work is path dependent since the steps taken to get from initial to final conditions can differ.
- Sun Jan 27, 2019 1:38 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Heat vs Enthalpy
- Replies: 4
- Views: 341
Re: Heat vs Enthalpy
Heat is a transfer of energy. Enthalpy is a state function and cannot be directly measured.
- Mon Jan 21, 2019 10:12 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Chem Eq. Module #4 Question 15 Conflicting Answers
- Replies: 2
- Views: 332
Re: Chem Eq. Module #4 Question 15 Conflicting Answers
The first explanation is correct. You are right in saying that water has no effect on the equilibrium constant. However, the question is asking about the equilibrium compositions, or concentrations. For this, I don't think you can assume water is in excess. Thus, when more water is added, the aqueou...
- Mon Jan 21, 2019 9:54 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Table 11.2
- Replies: 1
- Views: 211
Re: Table 11.2
I think K is referring to Kp, so it would just be Kp and Kc.
- Mon Jan 21, 2019 9:53 pm
- Forum: Identifying Acidic & Basic Salts
- Topic: Effect of pH
- Replies: 2
- Views: 405
Re: Effect of pH
This was in Chem14A, where some anions or cations are basic/acidic when dissociating in water so they can affect the pH. On page 478 in the 7th edition textbook there are a couple of tables that say which ions affect pH.
- Sun Jan 13, 2019 1:08 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Equilibrium Constant Eq.
- Replies: 4
- Views: 392
Re: Equilibrium Constant Eq.
No, since equilibrium constants are measuring concentrations or partial pressures, and the concentrations or pressure of solids and liquids are practically constant so they don't affect the concentration for ions or gases.
- Sun Jan 13, 2019 1:04 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Example in Class
- Replies: 7
- Views: 465
Re: Example in Class
By taking away the product, the reaction is no longer in equilibrium, so more product will be produced to replace the amount taken away. It's more product yield in the sense that more product is being made, not necessarily more total product.
- Sun Jan 13, 2019 12:55 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Changes in Partial Pressure
- Replies: 5
- Views: 440
Re: Changes in Partial Pressure
I think Dr. Lavelle said in class that pressure can also be changed by pumping an inert gas, so that wouldn't change the concentration compared to changing the volume, but K would still stay the same either way.
- Sat Dec 08, 2018 10:30 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Polydentates
- Replies: 2
- Views: 303
Re: Polydentates
You can tell if a molecule is polydentate if there are multiple atoms in a molecule that have lone pairs.
- Sat Dec 08, 2018 10:29 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: heisenberg's
- Replies: 5
- Views: 939
Re: heisenberg's
Yes, but generally the question will ask you to find the minimum value so you can just use the equal sign in that case.
- Sat Dec 08, 2018 10:27 pm
- Forum: Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)
- Topic: Hydrogen Bonding
- Replies: 5
- Views: 754
Hydrogen Bonding
I understand that hydrogen bonding occurs between hydrogen and N, O, F, but I'm confused why the hydrogen atoms on a molecule aren't counted as being able to form hydrogen bonds. Can't those hydrogen form hydrogen bonds with N, O, or F on another molecule? For example, a terminal oxygen atom is coun...
- Sun Dec 02, 2018 7:26 pm
- Forum: Properties & Structures of Inorganic & Organic Bases
- Topic: Ionization in Water
- Replies: 8
- Views: 1068
Re: Ionization in Water
Strong bases fully dissociate in water, while weak bases only partially dissociate, as the strength of the base depends on its ability to fully dissociate.
- Sun Dec 02, 2018 7:05 pm
- Forum: Lewis Acids & Bases
- Topic: Weak and Strong Acids/Bases
- Replies: 7
- Views: 785
Re: Weak and Strong Acids/Bases
For the acids, there are in total 7 strong acids that you should just memorize: HCl, HBr, HI, HNO3, HClO3, HClO4, H2SO4. (so the rest would be weak acids). The strong bases, instead you can recognize them as they are always a metal hydroxide (e.g. LiOH, NaOH, KOH...). For the metal hydroxides (stro...
- Sun Dec 02, 2018 6:49 pm
- Forum: Properties & Structures of Inorganic & Organic Bases
- Topic: Amines
- Replies: 1
- Views: 402
Re: Amines
Amines are basically any compound that has nitrogen with a lone pair, where it's derived from ammonia by replacing one or more of the hydrogen atoms.
For example, using R to stand in for the replacement groups, amines could be
NH3 ; NH2R ; NHR2
For example, using R to stand in for the replacement groups, amines could be
NH3 ; NH2R ; NHR2
- Sun Nov 25, 2018 5:17 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Xenon Tetrafluoride
- Replies: 3
- Views: 392
Re: Xenon Tetrafluoride
Xenon is in Period 5 which is after Period 3, so it can be an exception to the octet rule and have more regions of electron density, which allows it to have 6 regions of electron density for XeF4
- Sun Nov 25, 2018 5:06 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Hw: Ch.4 #73 (6th Edition)
- Replies: 2
- Views: 380
Re: Hw: Ch.4 #73 (6th Edition)
You can use the VSEPR model to help understand this. CH 2 2- has 2 bonding pairs (the 2 H atoms) and 2 lone pairs. The VSEPR formula then is AX 2 E 2 . Based on the formula, the molecular shape is bent. This is similar to the molecular geometry for water. The lone pairs exert greater repulsion compa...
- Sun Nov 25, 2018 5:00 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Polarity
- Replies: 3
- Views: 682
Re: Polarity
ClF3 has 2 lone pairs and its molecular shape is t-shaped. Because of the 2 lone pairs, the dipoles don't cancel, so ClF3 is polar.
- Sun Nov 18, 2018 8:43 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Lecture VSEPR
- Replies: 2
- Views: 373
Re: Lecture VSEPR
I think what it means is that VSEPR models can tell you the general shape of the molecule, such as bent, tetrahedral, square planar, etc, so that would be qualitative. Quantitative I think is the the angles of the bonds, so while VSEPR models can predict the general molecular geometry, it can't pred...
- Sun Nov 18, 2018 8:38 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Molecular Shape vs. Electron Geometry
- Replies: 4
- Views: 452
Re: Molecular Shape vs. Electron Geometry
Electron geometry is the geometry of all the electron groups, including lone pairs and bonds. Molecular shape is the actual position of the atoms, not including lone pairs. For example, NH 3 electron geometry is tetrahedral (due to 3 bonds and 1 lone pair), but the molecular geometry is trigonal pyr...
- Sun Nov 18, 2018 8:33 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Molecular Shape
- Replies: 4
- Views: 404
Re: Molecular Shape
I think that molecular shape is important in relation to biology, for example proteins get their function from their shape, so that's why the actual physical shape is more important than the electron geometry.
- Sat Nov 10, 2018 9:41 pm
- Forum: Polarisability of Anions, The Polarizing Power of Cations
- Topic: Polarizability Trend
- Replies: 6
- Views: 6519
Re: Polarizability Trend
Polarizability can be seen as trending the opposite way of electronegativity. Polarizability is how easy the electrons of an atom/molecule are distorted. If an atom has a high electronegativity, they have a stronger hold on their electrons, and thus the electrons are not easily distorted. If an atom...
- Sat Nov 10, 2018 9:30 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: how to know which shape it is
- Replies: 6
- Views: 704
Re: how to know which shape it is
The way to figure this out is using VSEPR. Basically you can use the Lewis structure to figure out the molecular geometry, using the VSEPR notation of AX n E y , where A represents the central atom, X represents the atoms bonded to the central atom, n is the number of bonds to the central atom (doub...
- Sat Nov 10, 2018 9:15 pm
- Forum: Bond Lengths & Energies
- Topic: 2D 15)
- Replies: 3
- Views: 735
Re: 2D 15)
Just remember that the shorter the bond, the stronger it is. Cl is bigger than F, so the bonds of C-Cl will be further apart than C-F, and so C-Cl bonds will be weaker than C-F. Br is bigger than Cl, so C-Br bonds will be further than C-Cl bonds, so C-Br will be weaker than C-Cl. C-F bonds are the s...
- Sat Nov 03, 2018 6:57 pm
- Forum: Quantum Numbers and The H-Atom
- Topic: Degeneracy
- Replies: 8
- Views: 1408
Re: Degeneracy
I believe that for elements with more than one electron, so basically every element except hydrogen, degeneracy is determined when the values for n and the values for l are the same.
- Sat Nov 03, 2018 5:49 pm
- Forum: Lewis Structures
- Topic: Lewis Structure of SO4^2- and Formal Charge
- Replies: 1
- Views: 478
Lewis Structure of SO4^2- and Formal Charge
In class professor Lavelle did an example where he drew the SO4^2- ion with all single bonds, and then drew another structure where there were 2 double bonds. Which one would be the correct way to draw it? Would both technically be correct, just that the second one is more correct due to the formal ...
- Sat Nov 03, 2018 5:22 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: Exam 2 Question 2
- Replies: 1
- Views: 309
Re: Exam 2 Question 2
I think you just forgot to solve for ∆p. You solved only for momentum (p) but just forgot the step to get ∆p. The question tells you that ∆p is (1x10^-6) times the momentum, you have to solve for p first using p = mv, then multiply the result by (1x10^-6). I think you just forgot that last part. Wha...
- Sun Oct 28, 2018 9:25 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Energy Release in Electron Affinity
- Replies: 1
- Views: 241
Energy Release in Electron Affinity
I know that electron affinity is the energy released to add an electron to an atom. I'm just a little confused on why energy is released? Why wouldn't energy be required to add an electron?
- Sun Oct 28, 2018 7:25 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: 7th edition 2A.9 and 2A.11
- Replies: 2
- Views: 379
Re: 7th edition 2A.9 and 2A.11
One way to figure the questions out is to just add up the number of electrons. For example, 2A.9 a) is [Ar]3d7. Ar has 18 electrons, plus the 7 given in 3d7, for a total of 25 electrons. The question tells you that it is a +2 charge ion, which means the neutral atom lost 2 electrons. 25 electrons is...
- Sun Oct 28, 2018 7:10 pm
- Forum: Ionic & Covalent Bonds
- Topic: Topic 2A.1 (valence electrons) 7th edition
- Replies: 3
- Views: 398
Re: Topic 2A.1 (valence electrons) 7th edition
Valence electrons are electrons that are in the outermost shell for an atom. The outermost shell for Sb then would be the 5p orbital, so the valence electrons would be the electrons in 5p, which is 5. It wouldn't include the d-orbital since the d-orbital is not the outermost shell.
- Sun Oct 21, 2018 9:08 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: h with a line through it?
- Replies: 8
- Views: 1941
Re: h with a line through it?
h bar is h/2π. The difference is in the Heisenberg equation where it could be (Δp x Δx) = h/4π or (Δp x Δx) = 1/2(h bar). 1/2(h bar) simplifies to h/4π
- Sun Oct 21, 2018 9:02 pm
- Forum: Einstein Equation
- Topic: Energy of photon [ENDORSED]
- Replies: 12
- Views: 1875
Re: Energy of photon [ENDORSED]
Henry_Phan_3B wrote:Chem_Mod wrote:Yes, E = hv or E = hc/.
How did you get that second equation?
You substitute for v in E=hv to get E=hc/λ
Use the light equation c = vλ, solve for v to get v = c/λ, and substitute back into E=hv for v
- Sun Oct 21, 2018 8:55 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Px. Py, Pz
- Replies: 6
- Views: 1708
Re: Px. Py, Pz
Is there a specific order for values to be assigned to Px, Py, Pz? In lecture I remember he said it was arbitrary and that it didn't matter? I'm a little confused on that.
- Sun Oct 14, 2018 9:55 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: HW #1.43
- Replies: 8
- Views: 1327
Re: HW #1.43
h bar is said in the textbook to be h/2π, where h is Planck's constant The uncertainty principle equation is (∆p)(∆x) = .5(h bar) or (∆p)(∆x) = h/4π ∆p is the momentum uncertainty and ∆x is the position uncertainty. ∆x is given in the problem, which is 350. pm or 3.50 x 10 -10 meters. Plug it into t...
- Sun Oct 14, 2018 9:31 pm
- Forum: Properties of Electrons
- Topic: Atomic Spectrum: Lines in a Series
- Replies: 2
- Views: 300
Re: Atomic Spectrum: Lines in a Series
The different series correspond to different parts of the hydrogen spectrum. For example, the Balmer series is used for the visible light spectrum of hydrogen, while the Lyman series is used for the ultraviolet spectrum of hydrogen. As Atul said, the series have different base energy levels. This is...
- Sun Oct 14, 2018 9:19 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: 6th edition 1.15
- Replies: 2
- Views: 146
Re: 6th edition 1.15
The way I did it was using the Rydberg equation, which is 1/wavelength (in meters) = R(1/n 1 2 - 1/n 2 2 ) The wavelength is given, and R is the Rydberg constant. Since this is in the ultraviolet spectrum of hydrogen, it is the Lyman series, where n 1 is 1. Now all you need to do is plug in and solv...
- Thu Oct 04, 2018 2:49 am
- Forum: Limiting Reactant Calculations
- Topic: Significant figures in textbook 7th edition [ENDORSED]
- Replies: 5
- Views: 381
Re: Significant figures in textbook 7th edition [ENDORSED]
I think the safest option is to just round to however many significant figures were given in the problem.
- Thu Oct 04, 2018 2:46 am
- Forum: Significant Figures
- Topic: M11 Sig Figs
- Replies: 4
- Views: 417
M11 Sig Figs
For part C on M11 (7th edition), the answer in the back is given only 2 sig figs as 5.7g. The answer I got was rounded to 3 sig figs, at 5.79g. I was just wondering it's supposed to be 3 sig figs, as the answer to part B is 3 sig figs, or if my answer is wrong? Since if I round it to 2 sig figs it w...
- Thu Oct 04, 2018 2:35 am
- Forum: Limiting Reactant Calculations
- Topic: M.11
- Replies: 8
- Views: 1608
Re: M.11
O2 is the limiting reactant for part a. You first have to figure out the limiting reactant for the first reaction using mole ratios, which turns out to be P4. This means that all of the P4 is used in the first reaction, but only some of the O2 is used. Since you now know that P4 is the limiting reac...