Search found 57 matches
- Sat Mar 17, 2018 8:16 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Catalysts
- Replies: 3
- Views: 573
Re: Catalysts
if it is seen in the overall reactant and also in the products therefore "cancelling out" and not being consumed
- Fri Mar 16, 2018 2:16 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: 8.103
- Replies: 3
- Views: 709
8.103
Are we supposed to know that average kinetic energy is equal to (3/2)RT? Overall, how does the solution manual obtain these answers and are we expected to know this formula even though it's not given on the sheet?
- Thu Mar 15, 2018 4:13 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Max work and delta G
- Replies: 1
- Views: 326
Max work and delta G
Conceptually, how does delta G = work max = -nFE ?
- Wed Mar 14, 2018 12:42 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Slow steps and intermediates
- Replies: 2
- Views: 325
Re: Slow steps and intermediates
No, that did not indicate that the second step was the slow step. You have to test it. I think he did that just to have us go through a little more work because if the first step was the slow step, it would have been an easier, shorter problem.
- Wed Mar 07, 2018 8:39 pm
- Forum: First Order Reactions
- Topic: UA worksheet [ENDORSED]
- Replies: 2
- Views: 372
Re: UA worksheet [ENDORSED]
If your half life is 6 days, in 30 days you will go through 5 half lives.
If you start with 100% you divide it by 2 (or multiply by 1/2) five times to get 3.1%
100/2=50% 1 half life
50/2=25% 2 half lives
25/2=12% 3 half lives
12/2=6.25% 4 half lives
6.25/2=3.125% 5 half lives
If you start with 100% you divide it by 2 (or multiply by 1/2) five times to get 3.1%
100/2=50% 1 half life
50/2=25% 2 half lives
25/2=12% 3 half lives
12/2=6.25% 4 half lives
6.25/2=3.125% 5 half lives
- Wed Mar 07, 2018 8:19 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Finding the intermediate(s)
- Replies: 5
- Views: 675
Re: Finding the intermediate(s)
If it is produced in a step and a product in another, therefore being cancelled out of the overall reaction, then it is an intermediate
- Wed Mar 07, 2018 10:21 am
- Forum: General Rate Laws
- Topic: Stoichiometric Coefficients
- Replies: 2
- Views: 340
Stoichiometric Coefficients
Why does unique rate use stoichiometric coefficients and the other equations do not need coefficients?
- Sun Mar 04, 2018 4:35 pm
- Forum: First Order Reactions
- Topic: Second Order Half Life Equation
- Replies: 2
- Views: 354
Re: Second Order Half Life Equation
.693 and my response is actually for a first order half life equation. t(1/2)= (ln2)/k
second order half life is t(1/2)=1/(k[A]0)
second order half life is t(1/2)=1/(k[A]0)
- Sun Mar 04, 2018 4:07 pm
- Forum: First Order Reactions
- Topic: Second Order Half Life Equation
- Replies: 2
- Views: 354
Re: Second Order Half Life Equation
second order half life numerator is always ln 2 because it is not dependent on initial concentration so .693 is just ln 2
- Sun Mar 04, 2018 3:52 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Stopped Flow Technique
- Replies: 3
- Views: 450
Re: Stopped Flow Technique
Reactions that are initiated by mixing the reagents can be studied by the stopped-flow technique, in which solutions of the reactants are forced into a mixing chamber very rapidly and, within a few milliseconds, the formation of products is observed spectroscopically. This procedure is commonly use...
- Sun Mar 04, 2018 3:39 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Stopped Flow Technique
- Replies: 3
- Views: 450
Stopped Flow Technique
What is the stopped flow technique and what is it's relation to kinetics?
- Sun Feb 25, 2018 1:04 pm
- Forum: First Order Reactions
- Topic: Pseudo Order
- Replies: 1
- Views: 251
Pseudo Order
What exactly is a pseudo order reaction? Do we need to know this?
- Wed Feb 21, 2018 9:37 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: 14.83 [ENDORSED]
- Replies: 2
- Views: 391
14.83 [ENDORSED]
Why is it that in part b, the solution manual says that E cell not and E cell are both temperature dependent? Shouldn't E cell not only be at 25 degrees Celsius? How is it temperature dependent then?
- Wed Feb 21, 2018 9:23 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Adiabatic System
- Replies: 3
- Views: 575
Re: Adiabatic System
it is thermally insulated. heat does not enter or leave the system
- Tue Feb 13, 2018 7:46 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: ΔG° vs ΔG
- Replies: 5
- Views: 4239
Re: ΔG° vs ΔG
ΔG° is Gibbs free energy at standard conditions. You use ΔG° to find ΔG by ΔG=ΔG°+ RTlnQ
- Tue Feb 13, 2018 7:44 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: PV = nRT
- Replies: 6
- Views: 940
Re: PV = nRT
When dealing with an ideal gas and if you are looking for one of the variables and given the others. You're not really directed to use it, but you use it to find a missing variable that you might need. For example, if you need to use volume but you're given moles, temperature, and pressure. Also, wo...
- Tue Feb 13, 2018 7:40 pm
- Forum: Calculating Work of Expansion
- Topic: -w vs +w
- Replies: 2
- Views: 386
Re: -w vs +w
If work is done by the system, -w, it means expansion. If work is done onto the system, +w, it means compression.
- Fri Feb 09, 2018 10:15 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Cv,m vs Cp,m vs R
- Replies: 4
- Views: 12766
Re: Cv,m vs Cp,m vs R
Also, if there are two changes in a combination of temperature, pressure, and volume, you would add the two steps together to get combined entropy change. In a review session, an example was entropy change of 2.00 mol of Argon gas volume from 10L to 5L and Temperature from 300K to 100K with Cv= 12.4...
- Fri Feb 09, 2018 10:12 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Cv,m vs Cp,m vs R
- Replies: 4
- Views: 12766
Re: Cv,m vs Cp,m vs R
Cv is specific heat in constant volume and Cp is specific heat in constant pressure. So when you are calculating entropy change with a change in temperature and constant volume, you would use the equation delta s = nCvln(T2/T1). If you are calculating entropy change with a change in temperature and ...
- Mon Feb 05, 2018 9:21 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 9.63
- Replies: 4
- Views: 460
9.63
Why is it that the compounds with a positive free energy of formation are unstable with respect to the elements?
- Sat Feb 03, 2018 3:01 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Extensive/Intensive Property [ENDORSED]
- Replies: 4
- Views: 576
Re: Extensive/Intensive Property [ENDORSED]
You can change an extensive property to intensive by dividing it by something like mass so that it is not "dependent" anymore
- Thu Feb 01, 2018 11:44 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: Third Law
- Replies: 3
- Views: 419
Re: Third Law
Yes the third law states that the limit of t approaching 0K of S will be equal to 0 for perfect monatomic crystals
- Wed Jan 31, 2018 2:28 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Chem Community Posts
- Replies: 5
- Views: 846
Re: Chem Community Posts
End of Sunday
- Tue Jan 23, 2018 11:25 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Bond Enthalpies
- Replies: 4
- Views: 431
Re: Bond Enthalpies
You use bond enthalpies to find the enthalpy of formation by subtracting the sum of the bond enthalpies of formed bonds from the sum of the bond enthalpies of the broken bonds.
H=sum of broken - sum of formed
H=sum of broken - sum of formed
- Tue Jan 23, 2018 11:23 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.73 and Units for Reaction Enthalpy
- Replies: 1
- Views: 197
Re: 8.73 and Units for Reaction Enthalpy
I think the book tends to denote enthalpy as kJ/mole of the reaction given, which does not mean necessarily per mole of the molecule. I believe my TA said that Dr. Lavelle usually doesn't write it this way in lecture as it can be confusing, but I believe the book just means kJ per mole of the reacti...
- Tue Jan 23, 2018 11:14 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Calorimeters Isolated
- Replies: 2
- Views: 539
Re: Calorimeters Isolated
The bomb calorimeter is isolated because it does not allow exchange of heat or matter, but the coffee cup calorimeter has to be open to have constant pressure.
- Fri Jan 19, 2018 4:09 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Concept Clarification
- Replies: 3
- Views: 282
Re: Concept Clarification
I would consider the ideal gas law PV=nRT because you are changing the number of moles which may change another variable, which would affect the energy of the system.
- Fri Jan 19, 2018 4:05 pm
- Forum: Calculating Work of Expansion
- Topic: 8.11
- Replies: 1
- Views: 208
8.11
How did the solutions manual go from -(1.00atm)(1.20L) to -122J?
- Fri Jan 19, 2018 3:10 pm
- Forum: Phase Changes & Related Calculations
- Topic: 8.47
- Replies: 3
- Views: 215
Re: 8.47
The work is positive because the work was done onto the system instead of by the system.
- Thu Jan 11, 2018 10:36 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.67
- Replies: 1
- Views: 1030
8.67
Can someone walk me through how to do this problem? (Use tables 8.3, 8.6, 8.7) Estimate the enthalpy of formation of each of the following compounds in the LIQUID state. (standard enthalpy of carbon sublimation is 717 kj/mol) a) H2O b) methanol, CH3OH c) benzene, C6H6 (without resonance) d) benzene,...
- Thu Jan 11, 2018 10:20 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Ch 8 Problem 49
- Replies: 1
- Views: 257
Re: Ch 8 Problem 49
For most equations we assume the temp is 25 degrees celsius, which is equivalent to 298K, so I think it was just assumed the equation took place at standard temperature since a temperature wasn't given.
- Thu Jan 11, 2018 10:16 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.61
- Replies: 4
- Views: 318
Re: 8.61
So you know the first equation is multiplied by 2 to get 2HBr, the second equation is reversed, and the third is left alone. If you write these out, it's easy to see which parts cancel. 2NH3, 2NH4Br, and N2 cancel leaving you with the desired H2+B2+2HBr. Remember to respectively change the delta H t...
- Fri Dec 08, 2017 10:36 am
- Forum: Conjugate Acids & Bases
- Topic: 12.3a
- Replies: 4
- Views: 646
Re: 12.3a
It would but in a two-step process. The first equation would yield HSO4- and the second would have HSO4- as the acid and yield SO4(2-)
- Fri Dec 08, 2017 10:34 am
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: 12.21
- Replies: 2
- Views: 309
Re: 12.21
It's an equilibrium constant at specifically 25 degrees celsius. Since it's a constant it's always the same number at this temperature.
- Sun Dec 03, 2017 12:23 pm
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: 12.35
- Replies: 1
- Views: 212
12.35
Give the Ka value for the acids
part d, HSeO4- pKa2=1.92
would you just solve for Ka2? Can you solve for Ka1? In the solution manual it has the answer for Ka2 but it is labeled Ka1? Or am I seeing it wrong?
part d, HSeO4- pKa2=1.92
would you just solve for Ka2? Can you solve for Ka1? In the solution manual it has the answer for Ka2 but it is labeled Ka1? Or am I seeing it wrong?
- Sun Dec 03, 2017 12:09 pm
- Forum: Lewis Acids & Bases
- Topic: Lewis Acid & Base vs. Bronsted Acid & Base [ENDORSED]
- Replies: 2
- Views: 342
Re: Lewis Acid & Base vs. Bronsted Acid & Base [ENDORSED]
Lewis acid: electron pair acceptor
Lewis base: electron pair donor
Bronsted acid: proton donor
Bronsted base: proton acceptor
(focus is just different!)
Lewis base: electron pair donor
Bronsted acid: proton donor
Bronsted base: proton acceptor
(focus is just different!)
- Thu Nov 23, 2017 11:49 am
- Forum: Naming
- Topic: H2O in coordination compounds
- Replies: 2
- Views: 395
H2O in coordination compounds
Does it matter if you write H2O or OH2 when writing the coordination compound if it has "aqua" in the name?
- Tue Nov 21, 2017 2:55 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: 11.41
- Replies: 1
- Views: 206
11.41
A 25.0-g sample of ammonium carbamate, NH4(NH2CO2) was placed in an evacuated 0.250-L flask and kept at 25 degrees C. At equilibrium, 17.4 mg of CO2 was present. What is the value of Kc for the decomposition of ammonium carbamate into ammonia and carbon dioxide? The rxn is NH4(NH2CO2) (s) <--> 2NH3 ...
- Mon Nov 13, 2017 11:46 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: "more or less than 109.5 degrees"?
- Replies: 4
- Views: 1910
Re: "more or less than 109.5 degrees"?
You are just supposed to understand that if you have molecules with lone pairs, it will "push" the other atoms to create an angle slightly less than what would usually be expected
- Mon Nov 13, 2017 11:42 pm
- Forum: Limiting Reactant Calculations
- Topic: Limiting Reactant vs Theoretical Yield
- Replies: 3
- Views: 1784
Re: Limiting Reactant vs Theoretical Yield
A limiting reactant is the reactant in an equation that determines how much product will be produced. This is found by determining how many moles of each reactant you have and comparing it to the ratio of reactants found when balancing the chemical equation. Theoretical yield is the maximum potentia...
- Tue Nov 07, 2017 5:29 pm
- Forum: Lewis Structures
- Topic: 3.97
- Replies: 2
- Views: 331
3.97
No one responded to me so I'm posting this question again
Is the solution manual indicating the shape of the molecule? Why is it drawn this way?
Is the solution manual indicating the shape of the molecule? Why is it drawn this way?
- Mon Nov 06, 2017 4:42 pm
- Forum: Dipole Moments
- Topic: Dipole Moment of CO
- Replies: 3
- Views: 3094
Re: Dipole Moment of CO
CO can be linear and have a dipole moment because it is not cancelling out. The oxygen will "pull" on the carbon creating a dipole moment. In comparison, CO2 does not have a dipole moment because it is linear and the bond dipoles cancel each other out.
- Mon Nov 06, 2017 4:38 pm
- Forum: Lewis Structures
- Topic: 3.97
- Replies: 2
- Views: 244
Re: 3.97
Chem_Mod wrote:Hi Kaylin,
Would you mind including a photo of said question and corresponding answer?
Thanks!
- Mon Nov 06, 2017 2:23 pm
- Forum: Lewis Structures
- Topic: 3.97
- Replies: 2
- Views: 244
3.97
Is the solution manual indicating the shape of the molecule? Why is it drawn this way?
- Fri Nov 03, 2017 11:01 pm
- Forum: Lewis Structures
- Topic: 3.67(b)
- Replies: 2
- Views: 384
3.67(b)
Why wouldn't ClO2 have double bonds instead of single bonds? (The solution manual shows the answer with single bonds)
- Fri Nov 03, 2017 10:50 pm
- Forum: Lewis Structures
- Topic: Extra electrons
- Replies: 1
- Views: 251
Extra electrons
In 3.61 for example, how do you know where to put the extra electrons? Iodine gets more lone pairs added instead of bonding more. Do you just try different structures and test for the lowest formal charge?
- Sat Oct 28, 2017 11:37 pm
- Forum: Lewis Structures
- Topic: Placement [ENDORSED]
- Replies: 6
- Views: 749
Re: Placement [ENDORSED]
The central atom tends to be the one with lower ionization energy or the least electronegativity, so you would have to know the trends of the periodic table.
- Sat Oct 28, 2017 11:03 pm
- Forum: Lewis Structures
- Topic: Central Atom [ENDORSED]
- Replies: 3
- Views: 432
Re: Central Atom [ENDORSED]
The atom with the least electronegativity or the lowest ionization energy tends to be the central atom.
- Sun Oct 22, 2017 4:47 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: unpaired electrons [ENDORSED]
- Replies: 4
- Views: 1301
unpaired electrons [ENDORSED]
How do you determine how many unpaired electrons an atom has? I understand the Aufbau principle and Hund's rule, so I'm wondering if you have to write out the electron configuration to figure this out or if there's an easier way?
- Sun Oct 22, 2017 3:03 pm
- Forum: Trends in The Periodic Table
- Topic: Atomic Size Trend [ENDORSED]
- Replies: 4
- Views: 500
Re: Atomic Size Trend [ENDORSED]
Very simply, as the number of protons increase, the added valence electrons will be pulled toward the nucleus more, which makes the radius smaller. So as you move right on the periodic table, more protons and electrons = greater attraction and smaller radius
- Sun Oct 15, 2017 8:15 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: Heisenberg's Uncertainty Principle - test
- Replies: 1
- Views: 244
Re: Heisenberg's Uncertainty Principle - test
Uncertainty is not on this test. It will cover up to de Broglie but you can read ahead to prepare for what's coming up since it is in the chapter.
- Sun Oct 15, 2017 8:13 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Why don't we use the Rydberg equation from the book?
- Replies: 8
- Views: 901
Re: Why don't we use the Rydberg equation from the book?
It does the same exact thing. Dr. Lavelle said he doesn't prefer the equation in the book because students were getting confused on which place to put the final and initial energy levels whereas the delta E equation is more clear and straight forward AND tells you which direction the energy is going...
- Sun Oct 15, 2017 8:08 pm
- Forum: DeBroglie Equation
- Topic: When to use DeBroglie equation? [ENDORSED]
- Replies: 4
- Views: 555
Re: When to use DeBroglie equation? [ENDORSED]
It describes how light has wavelike properties. In problems it is used when given momentum, mass, velocity, wavelength since those are the variables. You would not cross over this equation with the others though because that's about light as particles or photons.
- Sun Oct 08, 2017 5:01 pm
- Forum: SI Units, Unit Conversions
- Topic: Problem E7
- Replies: 3
- Views: 1333
Re: Problem E7
moles of Carbon = (2.1x10^9 atoms)/(6.0221415 × 10^23 mol^-1) = 3.5x10^-15 mol C
divide by Avogadro's number because it refers to how many atoms are in one mole
divide by Avogadro's number because it refers to how many atoms are in one mole
- Sun Oct 08, 2017 4:55 pm
- Forum: Limiting Reactant Calculations
- Topic: Calculating how much of the excess remains [ENDORSED]
- Replies: 4
- Views: 964
Re: Calculating how much of the excess remains [ENDORSED]
You would first find the limiting reactant and then use the ratio from the balanced chemical equation to find how much of the excess will be used. Then subtract how much will be used by the total excess... So if you have two moles of each reactant and the ratio is 1:2, the amount of excess used is 1...
- Sun Oct 08, 2017 4:48 pm
- Forum: Empirical & Molecular Formulas
- Topic: Friday Oct 6 Test
- Replies: 7
- Views: 1114
Re: Friday Oct 6 Test
^I believe that was the molecular and empirical formula since we were also given the molar mass
- Mon Oct 02, 2017 8:48 pm
- Forum: Molarity, Solutions, Dilutions
- Topic: G25 homework problem
- Replies: 7
- Views: 794
G25 homework problem
The question: Practitioners of the branch of alternative medicine known as homeopathy claim that very dilute solutions of substances can have an effect. Is the claim plausible? To explore this question, suppose that you prepare a solution of a supposedly active substance, X, with a molarity of 0.10 ...