Search found 59 matches
- Sun Mar 17, 2019 3:43 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: standard cell potential and free energy
- Replies: 3
- Views: 517
Re: standard cell potential and free energy
Standard Gibbs free energy is an extensive property because in the equation ∆G=-nFE, you have to multiply E by the number of electrons that are transferred.
- Sun Mar 17, 2019 3:40 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: rate constants at equilibrium
- Replies: 2
- Views: 611
Re: rate constants at equilibrium
K = 1 means that the products and reactants are at equal concentrations at equilibrium. Because this is not always the case, K does not always equal 1 at equilibrium.
- Tue Mar 12, 2019 11:21 pm
- Forum: First Order Reactions
- Topic: First Order Reactions
- Replies: 3
- Views: 442
First Order Reactions
Can 2A -> B + C be a first order reaction and why?
- Tue Mar 12, 2019 12:27 am
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: 6th ed 14.97
- Replies: 1
- Views: 282
6th ed 14.97
Use the data in Appendix 2B and the fact that, for the half- reaction F2(g) 2 H (aq) 2 e S 2 HF(aq), E 3.03 V, to calculate the value of Ka for HF.
How do you know the other half reaction for this cell?
How do you know the other half reaction for this cell?
- Tue Mar 12, 2019 12:03 am
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: 6th ed 14.55
- Replies: 1
- Views: 222
6th ed 14.55
14.55 A 1.0 m NiSO4(aq) solution was electrolyzed by using inert electrodes. Write (a) the cathode reaction; (b) the anode reaction. (c) With no overpotential or passivity at the electrodes, what is the minimum potential that must be supplied to the cell for the onset of electrolysis? For the anode ...
- Sun Mar 10, 2019 8:51 pm
- Forum: First Order Reactions
- Topic: Half life of 1st-Order Reaction
- Replies: 2
- Views: 372
Half life of 1st-Order Reaction
Conceptually, why does the half life of a first order reaction not depend on the initial concentration?
- Sun Mar 10, 2019 8:49 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Instantaneous vs Average
- Replies: 11
- Views: 1050
Re: Instantaneous vs Average
Average rate is the change in the value of a quantity divided by the elapsed time. The instantaneous rate on the other and is the rate of a change at an exact moment in time.
- Sun Mar 10, 2019 1:34 pm
- Forum: General Rate Laws
- Topic: Rate Determining Step
- Replies: 5
- Views: 2961
Rate Determining Step
Is the rate law for the rate determining step always equal to the overall rate law?
- Thu Feb 28, 2019 2:21 pm
- Forum: Administrative Questions and Class Announcements
- Topic: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)
- Replies: 179
- Views: 23763
Re: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)
For Worksheet 7 #8d, why is OH-(aq) not included in the cell diagram?
- Thu Feb 28, 2019 6:57 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: H+ and OH- in cell diagrams
- Replies: 1
- Views: 187
H+ and OH- in cell diagrams
Are we supposed to include H+ and OH- ions when we are drawing the cell diagram?
- Tue Feb 26, 2019 11:10 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Delta G at boiling point
- Replies: 2
- Views: 760
Delta G at boiling point
Why is standard Gibbs free energy for the vaporization of water at 100 degrees Celsius 0?
- Wed Feb 20, 2019 6:15 pm
- Forum: Accuracy, Precision, Mole, Other Definitions
- Topic: Hydrogen Molar Mass HW E9
- Replies: 5
- Views: 970
Re: Hydrogen Molar Mass HW E9
For the final answer, you would use the least number of sig figs that is reported in the question. For your example, I would plug in all 6 sig figs for that specific calculation, but my final answer would only have 3 sig figs.
I don't think that you have to use all 6 sig figs, this is just what I do.
I don't think that you have to use all 6 sig figs, this is just what I do.
- Wed Feb 20, 2019 6:09 pm
- Forum: Calculating Work of Expansion
- Topic: Work without change in volume
- Replies: 8
- Views: 1130
Re: Work without change in volume
If there is no change in volume, that means that the system has not expanded or been compressed. If there is no expansion or compression, there is no work because the work that we are referring to is only expansion/compression work.
- Wed Feb 20, 2019 6:07 pm
- Forum: Calculating Work of Expansion
- Topic: Expansion Against a Vacuum
- Replies: 4
- Views: 691
Re: Expansion Against a Vacuum
Yes, this would be an irreversible system
- Wed Feb 20, 2019 6:03 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Constant Pressure/Volume
- Replies: 6
- Views: 999
Re: Constant Pressure/Volume
If the reaction is taking place in an open beaker, pressure will be constant
- Wed Feb 20, 2019 6:01 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Entropy of Irreversible Expansion
- Replies: 2
- Views: 516
Re: Entropy of Irreversible Expansion
Entropy is a state function so the path that it takes from the initial to final state does not matter- only the initial and final states themselves. Therefore, it does not matter if the path was reversible or irreversible. If the initial and final states match, then the change in entropy will be the...
- Wed Feb 13, 2019 11:25 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Midterm
- Replies: 4
- Views: 572
Midterm
Will we have to know strong acids and bases for the midterm?
- Thu Feb 07, 2019 2:25 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Question 8.45
- Replies: 1
- Views: 195
Re: Question 8.45
Because there is 197 grams of Carbon, 197g/12.01g.mol^-1 = 16.4 moles of Carbon. 16.4 moles of carbon / 4 moles needed and multiply that by the standard enthalpy = (16.4 moles/4 moles)* 358.8 kJ = 1.47*10^3 kJ
- Wed Feb 06, 2019 6:47 pm
- Forum: Calculating Work of Expansion
- Topic: 8.9 6th edition
- Replies: 1
- Views: 172
8.9 6th edition
An ideal gas in a cylinder was placed in a heater and gained 5.50 kJ of energy as heat. If the cylinder increased in volume from 345 mL to 1846 mL against an atmospheric pressure of 750. Torr during this process, what is the change in internal energy of the gas in the cylinder? I found that the w = ...
- Wed Feb 06, 2019 6:34 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Problem 8.1 (c) 6th edition
- Replies: 2
- Views: 404
Re: Problem 8.1 (c) 6th edition
Coolant un a refrigerator coil is actually a closed system, not an isolated system.
- Sun Feb 03, 2019 9:22 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: CV,m and CP,m
- Replies: 1
- Views: 204
CV,m and CP,m
Which would you expect to be larger for a given substance, CV,m or CP,m and why?
- Sun Feb 03, 2019 7:40 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Specific Combustion Reaction
- Replies: 2
- Views: 301
Re: Specific Combustion Reaction
During a combustion reaction, a compound reacts with O2 and burns. The two reactants are H2 and O2. Because there is no Carbon in the reactants, CO2 would not form. Instead, the equation would be 2H2 + O2 --> 2H2O
- Sun Feb 03, 2019 7:36 pm
- Forum: Phase Changes & Related Calculations
- Topic: Water vapor vs boiling water burns
- Replies: 3
- Views: 480
Re: Water vapor vs boiling water burns
When you are burned by water vapor, the vapor has to go through a phase transition to turn from vapor to liquid. The vapor releases energy as it cools down to the liquid state. When you are burned by boiling water, it does not change phases. Because the water vapor has has a phase transition, it wil...
- Sun Feb 03, 2019 7:24 pm
- Forum: Phase Changes & Related Calculations
- Topic: Compression Work
- Replies: 3
- Views: 369
Re: Compression Work
If a piston is compressing, then work is being done to the system and the value of work would be positive.
- Thu Jan 24, 2019 10:54 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 7th edition 6D.5
- Replies: 3
- Views: 365
Re: 7th edition 6D.5
Given Ka, you can find the H+ concentration and the pH value. pOH = 14- pH and the concentration of OH- is 10^(-pOH).
Another way of finding the concentration of OH- is knowing that [OH-]*[H+]=10^-14. Once you have the concentration of H+, [OH-]=(10^-14)/[H+]
Another way of finding the concentration of OH- is knowing that [OH-]*[H+]=10^-14. Once you have the concentration of H+, [OH-]=(10^-14)/[H+]
- Thu Jan 24, 2019 10:48 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Test 1
- Replies: 8
- Views: 795
Test 1
Will we have to memorize strong acids and bases for Test 1?
- Thu Jan 24, 2019 10:47 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 11.89 (6th edition)
- Replies: 1
- Views: 154
Re: 11.89 (6th edition)
The graph gives you the initial and equilibrium partial pressures. A's partial pressure decreases by 10, B's partial pressure increases by 5, and C's partial pressure increases by 10. That means the molar ratio for A:B:C is 2:1:2 and the balanced equation is 2A ⇌ B + 2C
- Wed Jan 16, 2019 11:45 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 6th Edition 12.59
- Replies: 2
- Views: 524
6th Edition 12.59
Calculate the pH, pOH, and percentage protonation of solute in each of the following aqueous solutions: (a) 0.057 m NH3(aq); (b) 0.162 m NH2OH(aq); (c) 0.35 m (CH3)3N(aq); (d) 0.0073 m codeine, given that the pKa of its conjugate acid is 8.21. For parts b, c, and d, how are we supposed to determine ...
- Fri Jan 11, 2019 8:38 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 6th edition 11.43
- Replies: 1
- Views: 262
Re: 6th edition 11.43
For this question you would have to set up and ICE table. The initial composition for NO is given (1.0 bar), and the initial compositions for N2 and O2 would be 0 because the reaction hasn't taken place yet. X is the equilibrium concentration of N2 so the change in N2 would be +x. N2 and O2 have a 1...
- Fri Jan 11, 2019 8:28 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Problem 11.7
- Replies: 2
- Views: 141
Re: Problem 11.7
Yes, the flasks are shown in chronological order.
- Wed Jan 09, 2019 8:57 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Spectator Ions and Equilibrium Constants
- Replies: 2
- Views: 1340
Spectator Ions and Equilibrium Constants
For the reaction 2AgNO3(aq) + 2NaOH(aq) ⇌ Ag2O(s) + 2NaNO3(aq) + H2O(l), why is the equilibrium constant K = 1/[Ag+]^2[OH-]^2? What are spectator ions and why do they cancel?
- Sun Dec 09, 2018 8:43 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: A.X.E.
- Replies: 4
- Views: 651
Re: A.X.E.
A represents the central atom, X is the number of atoms bonded to the central atom, and E is the number of lone pairs on the central atom.
- Sun Dec 09, 2018 8:41 pm
- Forum: Administrative Questions and Class Announcements
- Topic: DOWNLOAD SESSION WORKSHEETS HERE - Sundays 4-6pm (Karen) [ENDORSED]
- Replies: 135
- Views: 39170
Re: DOWNLOAD SESSION WORKSHEETS HERE - Sundays 4-6pm (Karen) [ENDORSED]
For worksheet 9 #2, why is the answer BF3? How can Cl- act as a lewis base?
- Fri Dec 07, 2018 11:15 am
- Forum: Naming
- Topic: cyano vs cyanido
- Replies: 3
- Views: 444
cyano vs cyanido
What's the difference between cyano and cyanido? For example in Potassium tetracyanonickelate(II), why is it cyano and not cyanido?
- Sun Dec 02, 2018 11:52 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: VSEPR
- Replies: 4
- Views: 553
Re: VSEPR
If there is a central atom that the three atoms and the one lone pair is attached to, then the VSEPR formula would be AX3E. A is the central atom, X3 represents three atoms attached to the central atom, and E represents the one lone pair attached to the central atom.
- Sun Dec 02, 2018 11:49 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Polar Molecules
- Replies: 12
- Views: 933
Re: Polar Molecules
If the lewis structure is symmetrical, then it is non-polar because the dipole moments cancel out. If the lewis structure is asymmetrical or there are lone pairs on the central atom, then it is polar
- Sun Dec 02, 2018 11:46 pm
- Forum: Hybridization
- Topic: P orbital
- Replies: 7
- Views: 650
Re: P orbital
sp means that there are two regions of electron density. If there are three regions of electron density rather than two, then there would be another p, making it sp2
- Sun Nov 25, 2018 9:59 pm
- Forum: Hybridization
- Topic: Question 4.43 (Sixth Edition)
- Replies: 2
- Views: 268
Re: Question 4.43 (Sixth Edition)
In an sp3 orbital, it is 25% s-character and 75% p-character. In an sp2 orbital, it is 33% s-character and 67% p-character. Bond angle increases from sp3 to sp2 so bond angle increases as as the s-character of a hybrid orbital increases.
- Sun Nov 25, 2018 9:51 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Shape of CH2 (2-)
- Replies: 2
- Views: 284
Re: Shape of CH2 (2-)
The central atom, C, has four regions of electron density- two bonding pairs and two lone pairs. This makes a bent shape- similar to the H2O example that Dr. Lavelle went over in class.
- Sun Nov 25, 2018 9:46 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: HW 4.109
- Replies: 2
- Views: 171
Re: HW 4.109
The bond angle should be <109.5, but since it's only a little less than 109.5, it's still around 109.5
- Sun Nov 18, 2018 4:33 pm
- Forum: Bond Lengths & Energies
- Topic: 6th edition 3.87
- Replies: 2
- Views: 338
Re: 6th edition 3.87
In addition, the CF4 bond is the shortest because the size of an F is the smallest compared to Cl and Br. The smaller the size of the atoms, the shorter the bonds.
- Sun Nov 18, 2018 4:29 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: AX2E3 Shape [ENDORSED]
- Replies: 2
- Views: 257
AX2E3 Shape [ENDORSED]
Why is AX2E3 linear but AX2E2 bent?
- Sun Nov 18, 2018 4:21 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Water Molecule
- Replies: 3
- Views: 370
Water Molecule
For H2O, why are the two lone pairs next to each other? If lone pair- lone pair repulsion is the strongest, why aren't the two lone pairs on opposites sides of each other?
- Mon Nov 05, 2018 12:41 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Calculator for Midterm
- Replies: 2
- Views: 366
Calculator for Midterm
Can we bring a scientific calculator for the midterm?
- Sun Nov 04, 2018 12:11 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: Problem 1B.27
- Replies: 1
- Views: 166
Re: Problem 1B.27
The velocity is 5.00 +/- 5 so it would range from 0 (5-5) to 10 (5+5). The uncertainty in velocity would be 10-0=10.
- Sat Nov 03, 2018 10:10 pm
- Forum: DeBroglie Equation
- Topic: Hw problem 1.37- 6th edition
- Replies: 3
- Views: 698
Re: Hw problem 1.37- 6th edition
Both protons and neutrons have mass, so you would use the de Broglie equation. The mass of a proton is 1.6726*10^-27 kg and the mass of a neutron is 1.6749*10^-27 kg
- Fri Nov 02, 2018 5:25 pm
- Forum: Properties of Electrons
- Topic: Energy of an Electron
- Replies: 2
- Views: 613
Energy of an Electron
How do I find the energy of an electron if I'm only given its wavelength? Can I use the equation E=hv or does that only apply to particles without mass?
- Sun Oct 28, 2018 10:18 pm
- Forum: Trends in The Periodic Table
- Topic: Lewis Structures
- Replies: 7
- Views: 727
Re: Lewis Structures
The dots can go on any four sides, but keep in mind that two dots next to each other means that those two electrons are paired. If the 4 electrons that you're drawing aren't paired, the structure would have one dot on each of the 4 sides.
- Sun Oct 28, 2018 10:09 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: D and P orbital energies
- Replies: 3
- Views: 415
Re: D and P orbital energies
The 3d orbital has higher energy than the 4s orbital when the 4s orbital is not filled.
- Sun Oct 28, 2018 10:05 pm
- Forum: Properties of Light
- Topic: Speed of Light
- Replies: 12
- Views: 1009
Re: Speed of Light
No, the speed of light is the ultimate speed limit of the universe
- Sun Oct 21, 2018 7:50 pm
- Forum: DeBroglie Equation
- Topic: problem 43 6th edition
- Replies: 4
- Views: 433
Re: problem 43 6th edition
To solve for this, we would use the equation ∆x∆p = h/4π. For an electron, ∆p = m∆v, and therefore, m∆x∆v = h/4π. The uncertainty on position of an electron would be the diameter of the atom so ∆x = 350 pm. "m" is the mass of the electron which is 9.109*10^-31 kg and h=6.626*10^-34 J.s
- Sun Oct 21, 2018 7:20 pm
- Forum: Properties of Light
- Topic: measuring wavelength
- Replies: 5
- Views: 546
Re: measuring wavelength
In addition, I think Dr. Lavelle mentioned that anything less than 10^-18 m is not detectable.
- Sun Oct 21, 2018 1:15 pm
- Forum: Significant Figures
- Topic: Sig figs and percentages
- Replies: 3
- Views: 2247
Sig figs and percentages
Do percentages count as sig figs? For example, if we were to find 25% of 10.6, would the answer have 3 sig figs or 2?
- Fri Oct 12, 2018 11:27 am
- Forum: Molarity, Solutions, Dilutions
- Topic: E23 part b
- Replies: 4
- Views: 803
Re: E23 part b
For part b, it is asking for the amount in moles of SO3 molecules, so the answer would be in moles. Moles can be used to describe the amount of atoms, molecules, ions, electrons, other particles, or specified groups of such particles.
- Fri Oct 12, 2018 11:10 am
- Forum: SI Units, Unit Conversions
- Topic: mol vs. mmol
- Replies: 12
- Views: 2843
Re: mol vs. mmol
A mmol is a "millimole", and a "milli-" is a factor of 10^-3. So 1 mmol = 10^-3 mol.
- Fri Oct 12, 2018 11:02 am
- Forum: Properties of Light
- Topic: HW 1A.9 [ENDORSED]
- Replies: 4
- Views: 322
Re: HW 1A.9 [ENDORSED]
If you know the frequency (v), you can find the energy using the equation E=hv where h=6.626*10^-34 J.s. If you know the wavelength (λ), you can find the energy using the equation E=hc/λ where c=2.997*10^8m.s^-1.
- Thu Oct 04, 2018 9:54 pm
- Forum: Accuracy, Precision, Mole, Other Definitions
- Topic: Hydrogen Molar Mass HW E9
- Replies: 5
- Views: 970
Re: Hydrogen Molar Mass HW E9
I would use the value that is given on the periodic table that you are referring to. For our tests, we'll be given a periodic table and the answers will be based off of the sig figs on that periodic table.
- Thu Oct 04, 2018 9:39 pm
- Forum: Significant Figures
- Topic: Molar Mass
- Replies: 5
- Views: 451
Re: Molar Mass
Use the value that is given in the periodic table that you are referring to and then round the final answer to the correct number of sig figs.
- Thu Oct 04, 2018 9:34 pm
- Forum: Significant Figures
- Topic: Number of Sig Fig
- Replies: 8
- Views: 632
Re: Number of Sig Fig
The least number of sig figs in any number of the problem determines the number of significant figures in the answer. Trailing zeroes behind a decimal point will count as sig figs. For example, 6.48 / 2.16 = 3.00