Search found 61 matches
- Sun Mar 17, 2019 7:26 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Positive or Negative Ecell Values
- Replies: 4
- Views: 947
Re: Positive or Negative Ecell Values
Also, a positive E means a -delta G, which is a spontaneous reaction. This comes from the equation deltaG=-nFE
- Sun Mar 17, 2019 7:25 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Catalysts
- Replies: 4
- Views: 655
Re: Catalysts
Catalysts are not used up in the reaction. Rather, they just lower the activation energy of the reaction.
- Sun Mar 17, 2019 7:24 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Chemistry community posts
- Replies: 5
- Views: 986
Re: Chemistry community posts
He was pretty lenient last quarter and if you had missed a couple he still gave full points, so it should be the same.
- Sun Mar 10, 2019 8:47 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: What does K say about stability?
- Replies: 5
- Views: 3511
Re: What does K say about stability?
A large K means that there are more products than reactants at equilibrium. This means the products are more stable than the reactants, because reactions generally proceed to form the most stable things possible. Therefore, forming SO3 is more stable than forming SO2 and O2.
- Sun Mar 10, 2019 8:43 pm
- Forum: Zero Order Reactions
- Topic: Concentration independent of the rate
- Replies: 4
- Views: 521
Re: Concentration independent of the rate
Yes, zero order is independent of the concentration of reactants. This has a graph that has concentration as its y-axis, and time as the x-axis.
- Sun Mar 10, 2019 8:39 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Reaction Rate vs Rate of Consumption/Formation
- Replies: 3
- Views: 396
Re: Reaction Rate vs Rate of Consumption/Formation
The reaction rate or rate of reaction is the speed at which reactants are converted into products.
The formation rate is the rate that products are being formed.
The two definitions are very similar.
The formation rate is the rate that products are being formed.
The two definitions are very similar.
- Mon Mar 04, 2019 2:19 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Pt
- Replies: 14
- Views: 1243
Re: Pt
Basically, if there is no solid, you add Pt as your cathode or anode because it does not affect the reaction as it is an inert solid.
- Mon Mar 04, 2019 2:16 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: "N" in Kinetics
- Replies: 4
- Views: 480
Re: "N" in Kinetics
Don't forget that n=1 is usually assumed, so it might not be explicitly said.
- Mon Mar 04, 2019 2:14 pm
- Forum: First Order Reactions
- Topic: First Order Reactions
- Replies: 6
- Views: 573
Re: First Order Reactions
Yes, It will be a straight line. But I think Lavelle also said if you also see an exponential decrease, that would be [A] graphed and indicate a first order reaction as well
- Sun Feb 24, 2019 9:08 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Cell diagram
- Replies: 3
- Views: 330
Re: Cell diagram
To add on, a single vertical line represents an interface between phases - reactant|product. And a Salt Bridge consists of a gel containing a concentrated aqueous salt solution in an inverted U-tube. This allows the ions to flow without affecting the concentration and is denoted by double lines ॥
- Sun Feb 24, 2019 9:03 pm
- Forum: Balancing Redox Reactions
- Topic: Oxidation numbers
- Replies: 1
- Views: 173
Re: Oxidation numbers
The oxidation number for H is +1, so 6*(+1)=6.
The oxidation number for O is -2, so 1*(-2)=-2
Together that equals 4. Since there is no charge on the compound, set it equal to zero
2 (C) + 4 = 0
C=-4/2, C=-2
The oxidation number for O is -2, so 1*(-2)=-2
Together that equals 4. Since there is no charge on the compound, set it equal to zero
2 (C) + 4 = 0
C=-4/2, C=-2
- Sun Feb 24, 2019 9:00 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: gibbs and temperature, 9.67 6th ed
- Replies: 5
- Views: 587
Re: gibbs and temperature, 9.67 6th ed
In the seventh edition, it says to set delta G = to zero and solve for T to find the range of temperature using deltaG=delatH - TdeltaS. Any number from 0 to the found T value will be a spontaneous reaction.
- Fri Feb 15, 2019 7:38 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Equation based on equilibrium
- Replies: 5
- Views: 475
Re: Equation based on equilibrium
K indicates equilibrium, so you would use that equation for a reaction at equilibrium. You would use Q when it not.
- Fri Feb 15, 2019 7:36 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Gibbs Free Energy and Equilibrium
- Replies: 5
- Views: 508
Re: Gibbs Free Energy and Equilibrium
At zero, there is no more work that can be done, and the reaction does not favor either the reactants or products. Thus, it is at equilibrium.
- Fri Feb 15, 2019 7:34 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: delta G knot
- Replies: 4
- Views: 799
Re: delta G knot
delta G knot indicates standard conditions, while regular delta G is the change in free energy under other conditions
- Sun Feb 10, 2019 8:57 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Delta U for Reversible and Irreversible
- Replies: 4
- Views: 1402
Re: Delta U for Reversible and Irreversible
Also, deltaU = 0 for the isothermal reaction(or compression) of an ideal gas.
- Sun Feb 10, 2019 8:54 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: General entropy question
- Replies: 9
- Views: 813
Re: General entropy question
Entropy is, in simple terms, how many ways an object can position itself. Therefore, the higher the entropy is based on how complicated the molecule is. The more complicated, the more shapes it can form, and therefore the more entropy. This then means that entropy goes in the order gas>liquid>solid,...
- Sun Feb 10, 2019 8:50 pm
- Forum: Calculating Work of Expansion
- Topic: work equations
- Replies: 4
- Views: 452
Re: work equations
Basically - Isothermic, Reversible --> w=-nRTln(v2/v1)
Irreversible --> w = -P ΔV
Irreversible --> w = -P ΔV
- Sun Feb 03, 2019 3:08 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: What do do when moles are given
- Replies: 5
- Views: 470
Re: What do do when moles are given
You want to use the moles to find the concentration by dividing by the volume. The coefficients of the balanced equation give you your exponent.
- Sun Feb 03, 2019 3:04 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Gas vs solid
- Replies: 5
- Views: 495
Re: Gas vs solid
Since a solid is compact and the molecules do not move, it has less entropy. Since gas molecules are very spread out, they have more disorder and therefore more entropy.
- Sun Feb 03, 2019 3:02 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Heat VS. Energy
- Replies: 5
- Views: 7796
Re: Heat VS. Energy
Energy is the ability of a system to do work and the change of energy, while heat is the energy being transferred.
- Sun Jan 27, 2019 10:23 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: energy production
- Replies: 4
- Views: 503
Re: energy production
Glucose is used to form ATP, which is the actual stored energy that the heart uses. So breaking glucose bonds is not the direct source of energy.
- Sun Jan 27, 2019 10:21 am
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Adding Inert Gas
- Replies: 10
- Views: 4486
Re: Adding Inert Gas
An inert gas does not react with the species in the reactions. Therefore, it does not affect any of the concentrations, and K remains the same
- Sun Jan 27, 2019 10:19 am
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Changing Pressure
- Replies: 6
- Views: 685
Re: Changing Pressure
You want to look at which side has less moles of gas, once it is equilibrium and balanced. Increasing the pressure forces the reaction to decrease the amount of moles, therefore going towards whichever direction has the least amount of moles. And vice versa for decreasing pressure
- Sun Jan 20, 2019 7:05 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: endo/exothermic rxns
- Replies: 5
- Views: 448
Re: endo/exothermic rxns
Also, t the end of the equation it might give you delta H. If it is positive it is endothermic, while negative is exothermic.
- Sun Jan 20, 2019 7:03 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Bars vs. Molarity
- Replies: 5
- Views: 576
Re: Bars vs. Molarity
Also, if the question asks for Kc you know you need to use concentration
- Sun Jan 20, 2019 7:02 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Removing products
- Replies: 10
- Views: 3888
Re: Removing products
K does not change because it is a constant. If you remove products, the reaction would move to favor the production of more products to reestablish equilibrium. Therefore it would be Q that would change
- Sat Jan 12, 2019 8:58 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Calculating Kc
- Replies: 4
- Views: 454
Re: Calculating Kc
Blanking an equation gives you the correct stoichiometric values for the equation. This is important because you have to raise the concentration of a species to its stoichiometric concentration to get the correct Kc
- Sat Jan 12, 2019 8:56 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Kc and Kp
- Replies: 12
- Views: 1663
Re: Kc and Kp
Kp is only used for gases, so you won't be able to use liquids or solids for that reaction. But for reactions with liquids, solids, and aqueous, then no, you do not include liquids or solids. This is because a pure substance does not really have a concentration and therefore does not affect the equa...
- Sat Jan 12, 2019 8:54 am
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Endothermic
- Replies: 6
- Views: 587
Re: Endothermic
Endothermic reactions requires heat, and can basically be included as a reactant of the formation. Exothermic releases heat, and can be thought of as a product.
- Fri Dec 07, 2018 8:46 am
- Forum: Bronsted Acids & Bases
- Topic: Exceptions: HSO4- and H2PO4-
- Replies: 4
- Views: 1165
Re: Exceptions: HSO4- and H2PO4-
Since both anions have an H, they can donate it to a base. This means it can be acidic. I also believe that H2PO4- can donate two protons, but i'm not quite sure
- Fri Dec 07, 2018 8:40 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: Boiling Point
- Replies: 6
- Views: 2581
Re: Boiling Point
After identifying the intermolecular forces, you would want to look at the size, or amount of electrons in the atom. The more electrons there are, the more polarizable it is. The more polarizable element has the strongest Van der Waals forces and therefore Se would be the higher boiling point.
- Fri Dec 07, 2018 8:37 am
- Forum: Lewis Acids & Bases
- Topic: Determining which H^+ ions are given off
- Replies: 4
- Views: 326
Re: Determining which H^+ ions are given off
Generally it will be the protons that are farthest from the central atom. You also want to check stability of the Lewis Structure after the deprotonation.
- Sun Dec 02, 2018 9:05 pm
- Forum: Naming
- Topic: drawing ligands
- Replies: 6
- Views: 537
Re: drawing ligands
I think he wanted to show where the lone pairs on the atoms bonded together. They are the exact same, just one accurately shows the bonding.
- Sun Dec 02, 2018 9:03 pm
- Forum: Naming
- Topic: Knowing oxidation states
- Replies: 11
- Views: 903
Re: Knowing oxidation states
Yes, the charge is mandatory to know when naming compounds because you have to be able to make it neutral or charged, depending on what the question is asking
- Sun Dec 02, 2018 9:01 pm
- Forum: Hybridization
- Topic: P orbital
- Replies: 7
- Views: 607
Re: P orbital
To figure out the hybridization, you just count the number of electron density areas. Then, you equate it to the orbital levels. For example, if you have three electron density regions, you would write sp2 (s has 1 and p has 2). P can have up to three, and d can have up to 5
- Sun Nov 25, 2018 10:58 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Naming
- Replies: 4
- Views: 235
Re: Naming
I believe that would be stated directly in the question. If asked for molecular geometry, then you would give the VSEPR geometry which excludes the lone pair in naming the shape. If asked for electron domain geometry, then you would give the shape which includes the lone pairs
- Sun Nov 25, 2018 10:48 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: pi and sigma bonds
- Replies: 9
- Views: 1386
Re: pi and sigma bonds
A sigma bond is stronger because of the head to head overlapping, while a pi bond is weaker because of side to side overlapping. A sigma bond requires more energy to break than a pi bond does.
- Sun Nov 25, 2018 10:45 am
- Forum: Hybridization
- Topic: Electron Density vs Shape?
- Replies: 6
- Views: 666
Re: Electron Density vs Shape?
Electron density is given by all regions that have electrons, including both lone pairs and bonded atoms. However, the molecular geometry refers only to the atoms that are bonded to the central atom.
- Sun Nov 18, 2018 6:57 pm
- Forum: Dipole Moments
- Topic: Dipole moments in terms of very electronegative atoms
- Replies: 3
- Views: 449
Re: Dipole moments in terms of very electronegative atoms
Yes it would be because the dipole moments do not cancel out
- Sun Nov 18, 2018 6:54 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: No central atom
- Replies: 7
- Views: 588
Re: No central atom
I think you have to look at N as the central atom. So you would draw the Lewis Structure relative to each nitrogen, and bond them together. This would make it trigonal pyramidal
- Sun Nov 18, 2018 6:51 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Molecular Geometry
- Replies: 6
- Views: 595
Re: Molecular Geometry
Electron density geometry takes into account all the bonded regions between two atoms, as well as the lone pairs. However, molecular geometry is just the bonded regions of atoms in a molecule.
- Mon Nov 12, 2018 2:40 pm
- Forum: General Science Questions
- Topic: Chemistry Community Posts
- Replies: 8
- Views: 1113
Re: Chemistry Community Posts
How may should we have by now?
- Mon Nov 12, 2018 2:38 pm
- Forum: Properties of Light
- Topic: Test 3
- Replies: 24
- Views: 2110
Re: Test 3
Will we be tested on lewis structures again?
- Mon Nov 12, 2018 2:37 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 Bonds
- Replies: 14
- Views: 944
Re: Hydrogen Bonds
They bond with O,N,and F and are dipole-duple bonds
- Mon Nov 12, 2018 2:35 pm
- Forum: Formal Charge and Oxidation Numbers
- Topic: Formal Charge
- Replies: 12
- Views: 1214
Re: Formal Charge
It's obviously better if you do because of partial credit, but it's not necessary.
- Sun Nov 04, 2018 4:31 pm
- Forum: Coordinate Covalent Bonds
- Topic: Coordinate Covalent Bond Definition
- Replies: 14
- Views: 1799
Re: Coordinate Covalent Bond Definition
What would be an example problem that includes the concepts of coordinate covalent bonds?
- Sun Nov 04, 2018 4:27 pm
- Forum: Formal Charge and Oxidation Numbers
- Topic: Formal Charge
- Replies: 7
- Views: 936
Re: Formal Charge
Ions will have to have a formal charge, but that could be stated in the chemical formula given. For example, BrO- will have a formal charge of -1. There will be 6 lone pairs on Br, a covalent bond to O, with 6 lone pairs on O. This gives Br a formal charge of 0, and O a formal charge of -1. (FC = va...
- Sun Nov 04, 2018 4:21 pm
- Forum: Trends in The Periodic Table
- Topic: Ionization Energies
- Replies: 2
- Views: 253
Re: Ionization Energies
He said we wouldn't have to
- Sun Oct 28, 2018 6:21 pm
- Forum: Quantum Numbers and The H-Atom
- Topic: Degeneracy
- Replies: 5
- Views: 466
Re: Degeneracy
Is this the one where you would just square the value of (n) level to find the degeneracy?
- Sun Oct 28, 2018 6:20 pm
- Forum: Ionic & Covalent Bonds
- Topic: Trend of Ionic Radii
- Replies: 5
- Views: 499
Re: Trend of Ionic Radii
It decreases across a period (left to right) and increases down a group (top to bottom).
- Sun Oct 28, 2018 6:16 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: What are degenerate orbitals?
- Replies: 2
- Views: 333
Re: What are degenerate orbitals?
This was not discussed in lecture to the best of my memory either. Because degenerate orbitals are when there are two orbitals of the same energy, all you have to do is square the (n) level.
- Thu Oct 18, 2018 8:22 pm
- Forum: Properties of Light
- Topic: Test 2 [ENDORSED]
- Replies: 15
- Views: 1181
Re: Test 2 [ENDORSED]
Will he be testing on anything related to test 1?
- Thu Oct 18, 2018 8:19 pm
- Forum: Properties of Electrons
- Topic: eV to J
- Replies: 7
- Views: 873
Re: eV to J
So I get the conversion, but what exactly is an eV and what the difference between eV and J? Do they both measure energy?
- Thu Oct 18, 2018 8:17 pm
- Forum: Properties of Light
- Topic: Rydberg formula
- Replies: 8
- Views: 745
Re: Rydberg formula
I'm also confused. I thought Lavelle said we didn't have to know how to solve questions with Schroedinger's equation. Is it rydberg or schroedinger?
- Thu Oct 11, 2018 4:18 pm
- Forum: Properties of Electrons
- Topic: Atomic spectra
- Replies: 5
- Views: 426
Re: Atomic spectra
This spectroscopic analysis of light shows only photons of particular energy. By looking at the energy levels of the excited atoms, you can form the structure of your element and identify it.
- Thu Oct 11, 2018 4:15 pm
- Forum: Properties of Electrons
- Topic: Atomic Spectrum
- Replies: 2
- Views: 259
Re: Atomic Spectrum
The atomic spectra is the spectroscopic analysis of light given off by excited atoms, and shows only photons of a particular energy given off. The experiment takes the energy of the photon, and identifies its energy level. Then, using these energy levels, you can see the electronic structure of the ...
- Thu Oct 11, 2018 4:08 pm
- Forum: Properties of Light
- Topic: Why Short Wavelengths Can Eject e-
- Replies: 6
- Views: 918
Re: Why Short Wavelengths Can Eject e-
A photon must be able to reach the threshold energy required to eject an electron. Consider that the longer the wavelength, the lower the frequency, and the energy off each photon is proportional to its frequency, given by E=hv. Therefore, shorter the wavelength is, the more energy it has.
- Wed Oct 03, 2018 3:42 pm
- Forum: Limiting Reactant Calculations
- Topic: Finding theoretical yield
- Replies: 6
- Views: 633
Re: Finding theoretical yield
To add on, does this mean that the molar ration of one reactant to another is useless information in finding theoretical yield?
- Wed Oct 03, 2018 3:39 pm
- Forum: Balancing Chemical Reactions
- Topic: Combustion Reactions.
- Replies: 8
- Views: 2942
Re: Combustion Reactions.
To add onto this question, does it matter which order we put CO2 and H20 as the products? For example, can we write C6H6 + O2 --> H20 + CO2, or does it have to yield CO2 + H2O?
- Sun Sep 30, 2018 11:03 am
- Forum: Balancing Chemical Reactions
- Topic: Short cuts/methods for balancing equations
- Replies: 15
- Views: 1627
Re: Short cuts/methods for balancing equations
I have not been able to find any good youtube videos either. I find it easier to balance chemical equations by checking them one at a time from reactant to product and isolating them (I do it in my head but if it's too long I write it down). It's not bad to write it once or twice if its more complic...