Search found 110 matches
- Sun Mar 14, 2021 1:28 pm
- Forum: Student Social/Study Group
- Topic: How do you deal with burnout?
- Replies: 144
- Views: 13074
Re: How do you deal with burnout?
Taking enough restful breaks and listening to what your mind and body need are so important in this rough time. It's okay to be easier on yourself too, you have been working hard for many weeks now. :)
- Sun Mar 14, 2021 1:18 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Saying Thank You to Dr. Lavelle
- Replies: 490
- Views: 513040
Re: Saying Thank You to Dr. Lavelle
Thank you so much, Dr. Lavelle! I really appreciated how you cared for us students throughout the past few months in this difficult time. I'm grateful for how you helped us (especially first-years) feel connected to campus by recording all of your lectures in the lecture hall because I know that req...
- Thu Mar 11, 2021 11:23 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: ΔG°, ΔH°, and ΔS°
- Replies: 5
- Views: 1101
Re: ΔG°, ΔH°, and ΔS°
the ° symbol doesn't imply that the rxn is at 298 K, but rather that this is a value calculated from the reaction at equilibrium (i believe). so yes, we can calculate these values at different temperatures. i would assume that H and S will be adjusted to different temperatures but we're normally gi...
- Thu Mar 11, 2021 9:28 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: ΔG°, ΔH°, and ΔS°
- Replies: 5
- Views: 1101
ΔG°, ΔH°, and ΔS°
Does the ° symbol indicate that the temperature of the substance is at 298 K? I know that ° indicates that the substance is in the standard state (1 bar, 1 mol/L for a solute), but I was confused if ° means 298 K. I thought that ΔG° could vary depending on temperature, since ΔG° = ΔH° - TΔS°, meanin...
- Wed Mar 10, 2021 4:01 pm
- Forum: Zero Order Reactions
- Topic: zero order
- Replies: 47
- Views: 1764
Re: zero order
For a zero-order reaction, the rate of the reaction is independent of the concentration of the reactants. The differential rate law for zero-order is rate = (-1/a)(d[A]/dt) = k[A]^0 = k, so you can see how the rate does not depend on [A]. This can happen when a catalyst or enzyme is required in the ...
- Sun Mar 07, 2021 7:47 pm
- Forum: General Rate Laws
- Topic: Sapling HW Week 9/10 #13
- Replies: 3
- Views: 263
Re: Sapling HW Week 9/10 #13
Does anyone know why [OH-] can be included in the rate law? I thought [OH-] was also an intermediate.
- Sun Mar 07, 2021 7:45 pm
- Forum: First Order Reactions
- Topic: Textbook Problem 7B.3
- Replies: 2
- Views: 231
Re: Textbook Problem 7B.3
I first found the moles per L of A that would be used up in 115 s of this reaction. Since we know that 0.034 moles per L of B have formed, we can find the moles/L of A used up by using molar ratios based on the chemical equation given: (0.034 mol/L B)(2 mol A / 1 mol B) = 0.068 mol/L A. Now that we ...
- Sat Mar 06, 2021 11:10 pm
- Forum: Interesting Applications: Rechargeable Batteries (Cell Phones, Notebooks, Cars), Fuel Cells (Space Shuttle), Photovoltaic Cells (Solar Panels), Electrolysis, Rust
- Topic: Textbook Question 6.61
- Replies: 1
- Views: 180
Re: Textbook Question 6.61
The question asks what the potential difference (E) is between the inside and the outside of the cell only based on the K+ ions. The problem also includes information about the concentration of K+ ions inside vs. outside the cell, so from these two pieces of information, you can tell that K+ ions pa...
- Thu Mar 04, 2021 7:15 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Textbook 4I.9 Part B
- Replies: 1
- Views: 176
Re: Textbook 4I.9 Part B
For part b, isothermal means that ΔT = 0, so ΔU = (3/2) n R ΔT = 0. ΔU = q + w = 0 w = 0 because part b says free expansion, which means external pressure = 0. w = -P ex ΔV = 0. Since w = 0 and ΔU = 0, q must be equal to 0. This means that no heat is being transferred into the surroundings, so entro...
- Thu Mar 04, 2021 6:49 pm
- Forum: Balancing Redox Reactions
- Topic: Sapling week 7/8 Question 18
- Replies: 5
- Views: 386
Re: Sapling week 7/8 Question 18
I think the parentheses are implied but removed in Sapling's answer. So in total, there are still 2 x 3 H2O = 6 H2O molecules even without the parentheses being around Fe2O3∙3H2O.
- Sat Feb 27, 2021 2:23 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Textbook Problem 6M.3c [ENDORSED]
- Replies: 1
- Views: 157
Textbook Problem 6M.3c [ENDORSED]
Predict the standard potential of each of the following galvanic cells:
Hg (l) | Hg2Cl2 (s) | Cl− (aq) || Hg2 2+ (aq) | Hg (l)
Why is there no solid anode/cathode in this cell diagram? Is Hg a special case where its liquid form can be used as the anode/cathode?
Hg (l) | Hg2Cl2 (s) | Cl− (aq) || Hg2 2+ (aq) | Hg (l)
Why is there no solid anode/cathode in this cell diagram? Is Hg a special case where its liquid form can be used as the anode/cathode?
- Sat Feb 27, 2021 1:59 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Textbook Problem 6L.7c [ENDORSED]
- Replies: 1
- Views: 294
Textbook Problem 6L.7c [ENDORSED]
Write the half-reactions and devise a galvanic cell (write a cell diagram) to study each of the following reactions: Cd (s) + 2Ni(OH)3 (s) → Cd(OH)2 (s) + 2Ni(OH)2 (s), the reaction in the nickel−cadmium cell I was able to find the half-reactions: anode: Cd (s) + 2 OH- (aq) → Cd(OH)2 (s) + 2 e- cath...
- Sat Feb 27, 2021 11:55 am
- Forum: Balancing Redox Reactions
- Topic: Sapling week 7/8 Question 18
- Replies: 5
- Views: 386
Re: Sapling week 7/8 Question 18
I think your answer is correct, you just need to remove the parentheses around Fe2O3∙3H2O in order for Sapling to mark it correct.
- Sat Feb 27, 2021 11:49 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Textbook Problem 6L.3
- Replies: 1
- Views: 139
Re: Textbook Problem 6L.3
This post was helpful in explaining how to approach this problem: https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=139&p=325283&sid=ea29185e56002e4bd17838c9370b4edb#p325283 I also like to think of an anode / oxidation half-reaction as an oxidation number being increased. In this reaction,...
- Wed Feb 24, 2021 8:59 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Ecell vs E°cell
- Replies: 12
- Views: 1432
Re: Ecell vs E°cell
E°cell is the standard cell potential, or the cell potential at standard conditions (1 M solution, 1 atm, 298 K). Ecell is the cell potential at conditions that are not standard conditions.
- Wed Feb 24, 2021 6:06 pm
- Forum: Balancing Redox Reactions
- Topic: Sapling Week 7/8 #7
- Replies: 4
- Views: 301
Re: Sapling Week 7/8 #7
I can't seem to find anything wrong with your answer, I put the same answer as you and it was marked correct for me. Maybe check if you put in the right coefficients in each box?
- Sun Feb 21, 2021 5:19 pm
- Forum: Balancing Redox Reactions
- Topic: Week 7/8 Sapling #4
- Replies: 2
- Views: 142
Re: Week 7/8 Sapling #4
Yes, you would need to make sure the overall charge is equal on both sides of the equation. For Au(s), you know that the oxidation number is 0 because a neutral element that is not part of a compound has an oxidation number of 0. In this problem, H has an oxidation number of +1, and O has an oxidati...
- Sun Feb 21, 2021 4:57 pm
- Forum: Balancing Redox Reactions
- Topic: Sapling Week 7/8 #2
- Replies: 6
- Views: 306
Re: Sapling Week 7/8 #2
Besides adding electrons, you should also make sure that the number of electrons transferred is the same in both half-reactions. So, you would have to multiply both the reactions by a number to balance the number of electrons transferred: 2(In\rightarrow In^{3+} + 3 e^{-}) 3(Cd^{2+} + 2 ...
- Sun Feb 21, 2021 4:49 pm
- Forum: Balancing Redox Reactions
- Topic: Sapling HW 7/8 #1
- Replies: 8
- Views: 1356
Re: Sapling HW 7/8 #1
Is the best way to go about these types of problems memorizing the oxidation numbers for some elements? I don't think you necessarily need to memorize oxidation numbers for certain elements. Once you do more practice, you tend to pick up on the patterns. For example, oxygen tends to have an oxidati...
- Sun Feb 21, 2021 4:39 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Textbook 4.45
- Replies: 1
- Views: 162
Re: Textbook 4.45
The enthalpy of the solution is the ΔH of the system in the equation ΔS(surr) = - ΔH(sys) / T. Since ΔH(sys) = +34.9 kJ/mol, ΔS of the surroundings is a negative number. Only considering this decrease in entropy of the surroundings, we see that the dissolving process is nonspontaneous according to t...
- Thu Feb 18, 2021 11:58 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: textbook 4D.9
- Replies: 5
- Views: 651
Re: textbook 4D.9
I had the same question too ^ why does the enthalpy change per liter that we calculated become positive in the end?
- Thu Feb 18, 2021 11:31 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Textbook Question 4D.7
- Replies: 9
- Views: 492
Re: Textbook Question 4D.7
I was wondering this too, why do we assume that the temperature is 298 K for this problem?
- Sun Feb 14, 2021 8:19 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Textbook Problem 4F.1a
- Replies: 2
- Views: 147
Re: Textbook Problem 4F.1a
I agree with you, I was also was confused by the negative sign in the answer key. I think the rate of entropy generation should be positive, though, since the body is releasing heat into the surroundings, which would increase the entropy of the surroundings, so I think the answer should be +0.341 J/...
- Sun Feb 14, 2021 5:47 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Sapling week 5/6 #20
- Replies: 14
- Views: 2069
Re: Sapling week 5/6 #20
I think you are right, maybe try switching the answers so that spontaneous is in the top boxes, and Q<K / Q>K is in the bottom boxes. I had this same issue initially, but after switching them from top to bottom, it worked for me.
- Sun Feb 14, 2021 5:44 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: Textbook Problem 4H.11d
- Replies: 4
- Views: 238
Textbook Problem 4H.11d
Here is the question: Use the data in Table 4H.1 or Appendix 2A to calculate the standard reaction entropy for each of the following reactions at 25°C. For each reaction, interpret the sign and magnitude of the reaction entropy. (d) The decomposition of potassium chlorate: 4 KClO3 (s) → 3 KClO4 (s) ...
- Sun Feb 14, 2021 4:52 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Sapling Question#9
- Replies: 3
- Views: 204
Re: Sapling Question#9
To find ΔS of the surroundings, you can use the equation ΔS(surr) = -ΔH(sys) / T. ΔH of the system is −74.6 kJ (which you would have to convert to Joules), and T is 298 K. ΔS of the universe is the sum of ΔS of the system and ΔS of the surroundings. I hope this helps!
- Sun Feb 14, 2021 4:48 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Sapling #20
- Replies: 9
- Views: 420
Re: Sapling #20
When Q = K, the reaction is at equilibrium. This is indicated on the graph where the free energy (G) of the system is at its minimum. The free energy of the reactants is equal to the free energy of the products, so ΔG = 0.
- Sun Feb 07, 2021 3:37 pm
- Forum: Ideal Gases
- Topic: R ideal gas constant
- Replies: 31
- Views: 1956
Re: R ideal gas constant
I agree with everyone, there are different values of R because these values have different units, so you just need to make sure that you are using the R value that will leave you with the correct units in your answer.
- Sun Feb 07, 2021 3:30 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Textbook 4B.7
- Replies: 4
- Views: 133
Re: Textbook 4B.7
I think I understand your logic. Maybe the answer has a negative sign in order to emphasize the fact that the system (the fuel) is doing the work.
- Sun Feb 07, 2021 3:19 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Wks 3 & 4 Sapling Q18
- Replies: 6
- Views: 330
Wks 3 & 4 Sapling Q18
Here is the question: A 0.565 mol sample of CO2(g), initially at 298 K and 1.00 atm, is held at constant pressure while enough heat is applied to raise the temperature of the gas by 14.1 K. Calculate the amount of heat q required to bring about this temperature change, and find the corresponding tot...
- Sun Feb 07, 2021 3:10 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Sapling wk 3/4 #17
- Replies: 5
- Views: 539
Re: Sapling wk 3/4 #17
To find the specific heat of the gas, you can use the equation q = mcΔT, and you're solving for c, so you can rearrange the equation into c = q/(mΔT). However, we need to find q first using the equation ΔU = q + w, or q = ΔU - w. I hope this helps!
- Sun Feb 07, 2021 2:07 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Sapling Homework #20
- Replies: 6
- Views: 330
Re: Sapling Homework #20
To find the amount of heat (q), you can use the equation q = n C ΔT, where n = 0.201 mol and ΔT = 11.1 K. For C, you can use the molar heat capacity that is given in the problem, and I think the type of gas for this problem would be atoms, since the question involves Ar(g). ΔU = q + w, but we know t...
- Sun Jan 31, 2021 11:15 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: 6D.9 Textbook Question
- Replies: 2
- Views: 68
Re: 6D.9 Textbook Question
You don't need to know the formula in order to solve this problem. For example, you can just write the dissociation equation as: HA_{(aq)} + H_{2}O_{(l)} \rightleftharpoons H_{3}O^{+}_{(aq)} + A^{-}_{(aq)} Next, you can set up an ICE table. You know the initial concen...
- Sun Jan 31, 2021 11:02 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Sapling Week 3/4 #9
- Replies: 7
- Views: 264
Re: Sapling Week 3/4 #9
I agree with what everyone has said above! Once you set up the equation q of system = - q of surroundings, you can solve for Tf (final temperature). I considered the cold water to be the system and the warm water to be the surroundings.
- Sun Jan 31, 2021 12:37 pm
- Forum: Ideal Gases
- Topic: Reversing Reactions
- Replies: 68
- Views: 2238
Re: Reversing Reactions
Another way you can think of this is that when you reverse a reaction, you are multiplying the stoichiometric coefficients of the equation by -1, so the new K is K^-1. This is the same as how if you double the stoichiometric coefficients, the new K is K^2.
- Sun Jan 31, 2021 12:29 pm
- Forum: Calculating Work of Expansion
- Topic: why expanding volume has negative work
- Replies: 6
- Views: 352
Re: why expanding volume has negative work
I think the system has used energy in order to expand its volume, so the system has lost energy, which explains the negative sign.
- Sun Jan 31, 2021 12:23 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Change in Pressure/Compression on a system
- Replies: 7
- Views: 313
Re: Change in Pressure/Compression on a system
I agree with what everyone has said! I think compression does not significantly affect the concentration of liquids or aqueous solutions, so you would focus on comparing the moles of gas of reactants vs moles of gas of products in order to determine the shift in equilibrium.
- Thu Jan 21, 2021 7:30 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Textbook 6.B.3
- Replies: 3
- Views: 175
Re: Textbook 6.B.3
I agree with everyone! For another way to look at part b, the solution was diluted, so you can use M1 x V1 = M2 x V2. We're trying to find M2, the actual concentration of the solution prepared, and we can use this M2 to find the actual pH of the solution. M1 = 0.025 M H3O+, V1 = 0.2000 L, and V2 = 0...
- Thu Jan 21, 2021 7:22 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Week 2 Sapling #5
- Replies: 9
- Views: 248
Re: Week 2 Sapling #5
I agree, I did not disregard the X because the calculation is not too difficult even without disregarding the X. The quadratic formula is not needed in this calculation since we are solving for the initial concentration of the amine instead of solving for the equilibrium [OH-] like we usually do.
- Thu Jan 21, 2021 7:14 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: Textbook 6B.9
- Replies: 4
- Views: 205
Re: Textbook 6B.9
It just says "omit due to errors." Here's the solution manual errors: https://lavelle.chem.ucla.edu/wp-conten ... rs_7Ed.pdf
But I think as long as we understand the concept of this problem, which I'm sure you do, we'll be okay!
But I think as long as we understand the concept of this problem, which I'm sure you do, we'll be okay!
- Thu Jan 21, 2021 7:05 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: textbook 6.A.19
- Replies: 4
- Views: 220
Re: textbook 6.A.19
There's a typo in this question, so it should say 3.1 x 10^-3 mol.L^-1, not 3.1 mol.L^-1.
https://lavelle.chem.ucla.edu/wp-conten ... rs_7Ed.pdf
https://lavelle.chem.ucla.edu/wp-conten ... rs_7Ed.pdf
- Thu Jan 21, 2021 6:58 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 5.I.33
- Replies: 1
- Views: 78
Re: 5.I.33
This problem gives information that you can use to find the equilibrium concentration of CO2 (equilibrium mass and volume). Since ammonia carbamate is in the solid phase in this reaction, ammonia carbamate is not considered in the Kc value. The products NH3 and CO2 are both considered in Kc because ...
- Fri Jan 15, 2021 11:46 am
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Week 1 Sapling #9
- Replies: 5
- Views: 300
Re: Week 1 Sapling #9
For this question, you can first calculate Kc using the given equilibrium concentrations on the first line of the problem. You know that Kc will stay the same, since there is no temperature change mentioned in the problem. After the NO has been added, the system is no longer at equilibrium. So, you ...
- Thu Jan 14, 2021 7:04 pm
- Forum: Properties & Structures of Inorganic & Organic Acids
- Topic: Identifying Bases
- Replies: 7
- Views: 413
Re: Identifying Bases
Is there a quick way to identify a strong base from a weaker base? Like if it has an OH in it (for example, KOH) does that indicate anything as compared to a molecule without an OH (like CaO)? Strong bases are group 1 and group 2 oxides and hydroxides. For example, LiO2, NaOH, CaO, and Mg(OH)2 are ...
- Thu Jan 14, 2021 6:58 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Long term vs short term changes in conc.
- Replies: 6
- Views: 287
Re: Long term vs short term changes in conc.
I think you are correct! Adding a reactant results in an increase in that reactant after a short period of time. But in the long term, the reaction proceeds towards the formation of the products, meaning that there is a decrease in that reactant.
- Thu Jan 14, 2021 6:50 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: pKa and pH
- Replies: 10
- Views: 545
Re: pKa and pH
To add on to what everyone else has said, pKa is a fixed value for a given acid. pKa allows us to compare the strength of different acids, no matter what the concentration of each acid is, since a lower pKa value means that the acid is stronger. pH depends on the H3O+ concentration. This means that ...
- Mon Jan 11, 2021 10:45 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Textbook Problem 5.39
- Replies: 1
- Views: 135
Re: Textbook Problem 5.39
Table 5G.2 had an error with the given Kc value. It should be 6.1 x 10^-3 instead of 6.1 x 10^23.
Dr. Lavelle updated the solutions manual errors: https://lavelle.chem.ucla.edu/wp-conten ... rs_7Ed.pdf
Dr. Lavelle updated the solutions manual errors: https://lavelle.chem.ucla.edu/wp-conten ... rs_7Ed.pdf
- Mon Jan 11, 2021 10:42 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Textbook Problem 5.39
- Replies: 6
- Views: 339
Re: Textbook Problem 5.39
Edward Tang 1k wrote:Dr.Lavelle has updated the document solution manual errors. I've copied the part pertaining to this question below:
Table 5G.2 in the textbook has an error.
For the reaction N2O4 ⇌ 2NO2 at 298 K, the Kc value is 6.1 x 10-3
Thank you for the update! :)
- Sat Jan 09, 2021 12:43 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Textbook Problem 5.39
- Replies: 6
- Views: 339
Re: Textbook Problem 5.39
I think there might be a typo or error with this problem. When I checked the solutions manual, the Kc value (for ) used in the calculation was 6.1 x 10^-3, but Table 5G.2 says the Kc is 6.1 x 10^23, so I'm not sure which Kc value we should use.
- Thu Jan 07, 2021 4:20 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Preset values of K
- Replies: 5
- Views: 143
Re: Preset values of K
I agree with Hannah, I think the equilibrium constant is only dependent on temperature. Even with a pressure change, the system would respond to this change until the system reaches equilibrium again, and the equilibrium constant still remains the same.
- Thu Jan 07, 2021 4:16 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Chemical Equilibrium Part 3, #20
- Replies: 4
- Views: 141
Re: Chemical Equilibrium Part 3, #20
I agree with Hannah, for my x-value I also got 4.94 x 10^-6 because the numerator was (2x)^2.
- Wed Jan 06, 2021 12:38 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Converting bar to mol/L
- Replies: 5
- Views: 3109
Re: Converting bar to mol/L
Here's the unit conversion: \frac{nRT}{V}=P (\frac{mol}{L})(\frac{L\bullet atm}{K\bullet mol})(K)=atm (\frac{L\bullet atm}{K\bullet mol}) is the units for the constant R. In the lecture today, Dr. Lavelle said 1 bar is equal to approximately 1 atm, so we are usually g...
- Wed Jan 06, 2021 12:30 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Solvents absence in equilibrium constant eq
- Replies: 3
- Views: 224
Re: Solvents absence in equilibrium constant eq
To add on, when writing the Kc expression, even if you included the concentration of solvent as a product in the numerator and the concentration of solvent as a reactant in the denominator, these concentrations would be so similar in value that they would end up canceling (dividing them would give a...
- Wed Jan 06, 2021 12:21 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Part 3 Module Review Question 16
- Replies: 2
- Views: 379
Re: Part 3 Module Review Question 16
Since the initial concentration of PCl5 and Kc are given, I set up an ICE (Initial concentration, Change in concentration, Equilibrium concentration) table to find the equilibrium concentrations. Initial concentrations: 1.50 mol / 0.500 L = 3.00 M PCl5 ; 0 M PCl3 ; 0 M Cl2 Change in concentration: -...
- Tue Dec 08, 2020 1:56 pm
- Forum: Bond Lengths & Energies
- Topic: Compound stability
- Replies: 5
- Views: 350
Re: Compound stability
I think we look at the intramolecular bonds when determining the most stable compound. IMFs can be taken into account when we think about the solid, liquid or gas phase of a compound at room temperature, or when we think about melting or boiling points.
- Tue Dec 08, 2020 1:51 pm
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: Textbook question 6C.17
- Replies: 3
- Views: 218
Re: Textbook question 6C.17
I found the pKb values in the textbook if you scroll up in this section.
The pKb of morphine is 5.79 (from Table 6C.2)
The pKa of HBrO is given as 8.69 (from Table 6C.1). To find the pKb of BrO-, I subtracted 14.00 - 8.69 = 5.31.
The pKb of morphine is 5.79 (from Table 6C.2)
The pKa of HBrO is given as 8.69 (from Table 6C.1). To find the pKb of BrO-, I subtracted 14.00 - 8.69 = 5.31.
- Mon Dec 07, 2020 9:00 pm
- Forum: Hybridization
- Topic: Textbook Problem 2F.17
- Replies: 2
- Views: 161
Re: Textbook Problem 2F.17
O also has a hybridization of sp2 because the O atom has 3 regions of electron density: 1 (double) bond with the central C atom + 2 lone pairs. 3 regions of electron density gives the hybridization of sp2. I hope this helps!
- Mon Dec 07, 2020 8:29 pm
- Forum: Conjugate Acids & Bases
- Topic: acidic and basic salts
- Replies: 2
- Views: 160
Re: acidic and basic salts
A salt is basic when the cation is from group 1 or 2 on the periodic table, and the anion is the conjugate base of a weak acid. For example, in the salt NaF, the F- ion is the conjugate base of the weak acid HF (F- will remove a proton from water, generating OH-). Na+ does not affect the pH because ...
- Mon Dec 07, 2020 2:55 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Determining Valence e-
- Replies: 3
- Views: 203
Re: Determining Valence e-
I'm not quite entirely sure either, but here is what I think. For Mn for example, since its electron configuration is [Ar] 3d5 4s2, I would think Mn has 7 valence electrons (5 in the 3d subshell and 2 in the 4s subshell). When Mn becomes Mn 4+, it thus loses 4 valence electrons, 2 electrons from the...
- Mon Nov 30, 2020 9:57 pm
- Forum: Hybridization
- Topic: pcl3
- Replies: 8
- Views: 756
Re: pcl3
The lone pair on the central P atom and the three P-Cl bonds all count as regions of electron density, so there are 4 total regions of electron density. Since the number of regions of electron density = the number of hybrid orbitals, we know there must be 4 sp3 hybridized orbitals to represent the V...
- Mon Nov 30, 2020 9:52 pm
- Forum: Biological Examples
- Topic: Hemoglobin Subunits
- Replies: 3
- Views: 309
Re: Hemoglobin Subunits
I agree with Brittney, I think myoglobin refers to the entire molecule, including the Heme complex and histidine. For hemoglobin, I think we just have to know that hemoglobin has four myoglobin-like molecules, but the structure of each subunit of hemoglobin is not entirely the same as myoglobin.
- Mon Nov 30, 2020 9:46 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: AX2E3 Bond Angle
- Replies: 7
- Views: 513
Re: AX2E3 Bond Angle
Thank you all!
- Mon Nov 30, 2020 9:42 pm
- Forum: Hybridization
- Topic: identification
- Replies: 5
- Views: 314
Re: identification
First, you can determine the number of regions of electron density in order to determine the number of hybrid orbitals in BrF3 (number of regions of electron density = number of hybrid orbitals). After drawing the Lewis structure, we see that there are 5 regions of electron density, meaning there mu...
- Mon Nov 30, 2020 9:36 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Week 7 & 8 Sapling #18
- Replies: 2
- Views: 69
Re: Week 7 & 8 Sapling #18
Here's a picture of H2CCCH2 that hopefully helps in visualizing the orbitals. The central C atom has 2 sp2 hybridized orbitals (in blue) and 2 unhybridized p-orbitals (in green). These 2 p-orbitals (involved in the 2 π bonds) must be perpendicular to one another (like how px, py, and pz are all perp...
- Mon Nov 30, 2020 3:33 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: AX2E3 Bond Angle
- Replies: 7
- Views: 513
AX2E3 Bond Angle
Since the shape of AX2E3 is linear, would its bond angle be 180 degrees?
- Tue Nov 24, 2020 5:51 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Sapling #6
- Replies: 5
- Views: 215
Re: Sapling #6
Here is a picture I found online of the shape of XeF2. The lone pairs are represented by the color yellow, and the F atoms are represented by white. I hope this helps!
- Tue Nov 24, 2020 5:44 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Chem 14B
- Replies: 9
- Views: 429
Re: Chem 14B
I think he means that we might take our exams during lecture time instead of during our discussion sections, so we should keep our lecture times open just in case.
- Mon Nov 23, 2020 7:20 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Nonpolar molecules
- Replies: 3
- Views: 326
Re: Nonpolar molecules
A molecule is nonpolar when the molecule has no net dipole moment, such as when the the dipole moments in a molecule cancel each other. For example, CF4 is a nonpolar molecule. The difference in electronegativity between C and F creates a dipole moment. These four dipole moments cancel each other ou...
- Mon Nov 23, 2020 7:04 pm
- Forum: Hybridization
- Topic: hybridization of phosphorus (Sapling Q.11)
- Replies: 21
- Views: 3266
Re: hybridization of phosphorus (Sapling Q.11)
Looking at the Lewis structure, you see that the P atom has four regions of electron density. This means that there should be four hybrid orbitals. Even though there is a double bond, this still counts as one region of electron density. This might make a bit more sense if we think about the molecule...
- Mon Nov 23, 2020 6:58 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Wednesday Sections
- Replies: 1
- Views: 71
Re: Wednesday Sections
Yes, there will be discussion section and lecture on Wednesday.
- Thu Nov 19, 2020 4:49 pm
- Forum: Ionic & Covalent Bonds
- Topic: Ionic v. Covalent Character
- Replies: 6
- Views: 288
Re: Ionic v. Covalent Character
In an ionic bond, you would expect a complete transfer of an electron from one atom to another when the atoms are bonded together. However, we know that ionic bonds have some covalent character (covalent bonds entail sharing of electrons). This covalent character in ionic bonds is due to that shared...
- Thu Nov 19, 2020 4:43 pm
- Forum: Bond Lengths & Energies
- Topic: Textbook Question 3F.15
- Replies: 8
- Views: 466
Re: Textbook Question 3F.15
To add on to the previous responses, when you draw the Lewis structure of AsF3, you will notice that As, the central atom, has two lone pair electrons. This lone pair is the reason why AsF3 is a polar molecule. The lone pair creates a net dipole moment in the molecule, so we know that AsF3 has dipol...
- Thu Nov 19, 2020 4:36 pm
- Forum: Bond Lengths & Energies
- Topic: Bond Strength and Polarizability
- Replies: 4
- Views: 304
Re: Bond Strength and Polarizability
I think you might be confusing intramolecular and intermolecular forces as you said. Even if a molecule has strong bonds (which are intramolecular), it doesn't necessarily mean it will have strong intermolecular forces (such as London forces). I hope this helps!
- Wed Nov 18, 2020 5:38 pm
- Forum: Lewis Structures
- Topic: Textbook Problem 2C.9
- Replies: 1
- Views: 95
Re: Textbook Problem 2C.9
Before drawing the Lewis structure, you should count the total number of electrons in each molecule. For example, for XeF2, Xe has 8 electrons, and 7 electrons for F x 2 F atoms = 14 electrons. XeF2 has a total of 8 + 14 = 22 electrons. Next, you connect each F atom to the central atom Xe. With thes...
- Wed Nov 18, 2020 4:57 pm
- Forum: Octet Exceptions
- Topic: Textbook 2C.5
- Replies: 3
- Views: 183
Re: Textbook 2C.5
I agree with Sophia, it might have something to do with O having a higher electronegativity than Cl, meaning that O has a stronger pull on electrons than Cl does.
- Fri Nov 13, 2020 6:40 pm
- Forum: Lewis Acids & Bases
- Topic: Determining Lewis Acids and Bases
- Replies: 9
- Views: 524
Re: Determining Lewis Acids and Bases
To add on, molecules that have electron-deficient central atoms (for example, BeCl2) accept electrons and are thus Lewis acids.
- Fri Nov 13, 2020 6:31 pm
- Forum: Dipole Moments
- Topic: Sapling W 5/6 #17
- Replies: 10
- Views: 479
Re: Sapling W 5/6 #17
To add on, BCl3 also only has dispersion forces. For BCl3, Br2, and C2H6, the dipoles cancel out in each molecule, so there is no net dipole moment in each molecule, which tells us that there are no dipole-dipole interactions, only dispersion forces. CH3Cl has dipole-dipole interactions because the ...
- Fri Nov 13, 2020 6:12 pm
- Forum: Lewis Structures
- Topic: Sapling Week 5-6 HW Question 6
- Replies: 7
- Views: 306
Re: Sapling Week 5-6 HW Question 6
I am having some trouble with this question as well. In the explanation for why CO2 is a Lewis acid, Sapling says that molecules with polar double bonds accept electrons, but why are the double bonds in CO2 polar? Oxygen has a far greater electronegativity compared to Carbon, so the covalent bond i...
- Tue Nov 10, 2020 5:16 pm
- Forum: Dipole Moments
- Topic: H-bond and Dipole-Dipole
- Replies: 4
- Views: 191
Re: H-bond and Dipole-Dipole
I think a hydrogen bond is a special type of dipole-dipole interaction. It's kind of like how a square is always rectangle, but a rectangle isn't always a square. Hydrogen bonding is an intermolecular interaction that must involve two dipoles, but dipole-dipole interactions aren't always considered ...
- Tue Nov 10, 2020 5:08 pm
- Forum: Dipole Moments
- Topic: Sapling Q14
- Replies: 2
- Views: 91
Re: Sapling Q14
Claire_Latendresse_3J wrote:Hi Isabella,
Your reasoning is correct :) An N-H, O-H, or (sometimes) F-H bond can form hydrogen bonds, but a C-H bond has too small of an electronegativity difference to be viable for a hydrogen bond.
Thank you, that makes sense!!
- Tue Nov 10, 2020 4:56 pm
- Forum: Dipole Moments
- Topic: Sapling Q14
- Replies: 2
- Views: 91
Sapling Q14
Here is the question: Acetone can form hydrogen bonds with water. Four images are given which show an acetone molecule interacting with a water molecule. Hydrogen bonds are represented as dashed, green lines. I was wondering if my reasoning for this problem was correct. In this image, is the depicte...
- Tue Nov 10, 2020 11:33 am
- Forum: Lewis Structures
- Topic: Sapling Week 5-6 HW Question 6
- Replies: 7
- Views: 306
Re: Sapling Week 5-6 HW Question 6
I am having some trouble with this question as well. In the explanation for why CO2 is a Lewis acid, Sapling says that molecules with polar double bonds accept electrons, but why are the double bonds in CO2 polar?
- Wed Nov 04, 2020 9:51 pm
- Forum: Ionic & Covalent Bonds
- Topic: Textbook Problem 2A.1
- Replies: 4
- Views: 115
Re: Textbook Problem 2A.1
lauren_tran_3J wrote:I believe that even though d is a lower energy level it still has them as valence electrons because the shell is incomplete. I might be wrong! Hope this helps.
I was thinking about this too! Thank you for your help.
- Wed Nov 04, 2020 9:48 pm
- Forum: Octet Exceptions
- Topic: Incomplete Octet example BF3
- Replies: 1
- Views: 135
Re: Incomplete Octet example BF3
I would think that the top structure is the best Lewis structure when taking formal charge into consideration, since the formal charge of each atom is 0. The bottom structure with the double bond does not seem favorable to me because the formal charge of B is -1, and the formal charge of F is +1. Ma...
- Wed Nov 04, 2020 9:39 pm
- Forum: Octet Exceptions
- Topic: "Octets" beyond 8
- Replies: 6
- Views: 278
Re: "Octets" beyond 8
I don't think an expanded octet necessarily means the atom will be unstable. In the lecture, Dr. Lavelle said that certain atoms can accommodate more than 8 valence electrons because atoms in period 3 or higher have d-orbitals in their valence shell that can accommodate additional electrons.
- Wed Nov 04, 2020 9:31 pm
- Forum: Trends in The Periodic Table
- Topic: Effective nuclear charge
- Replies: 5
- Views: 1100
Re: Effective nuclear charge
Effective nuclear charge is the net nuclear charge after taking into account the shielding caused by other electrons in the atom. Inner electrons shield outer electrons from the electrostatic attraction of the positive nucleus, so the effective nuclear charge is less than the actual nuclear charge. ...
- Wed Nov 04, 2020 4:33 pm
- Forum: Ionic & Covalent Bonds
- Topic: Textbook Problem 2A.1
- Replies: 4
- Views: 115
Textbook Problem 2A.1
Here is the question: Give the number of valence electrons (including d electrons) for each of the following elements: (a) Sb; (b) Si; (c) Mn; (d) B. I am confused as to which electrons count as valence electrons, specifically for Sb and Mn. Sb has 5 valence electrons, while Mn has 7 valence electro...
- Thu Oct 29, 2020 9:12 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Bohr frequency question
- Replies: 2
- Views: 156
Re: Bohr frequency question
Yes, when the electron drops to a lower energy level, the energy difference is given off as a photon. This release of energy explains the negative sign, like you said. Also, like Kayla said, according to the reference point, a bound electron has lower energy than a free electron.
- Thu Oct 29, 2020 9:04 pm
- Forum: Trends in The Periodic Table
- Topic: Oxygen ionization energy
- Replies: 3
- Views: 198
Re: Oxygen ionization energy
The ionization energy of oxygen is an exception to the general trend in the periodic table. For oxygen, there are four electrons in the 2p subshell. Two of these electrons are in the same orbital, so they experience electron-electron repulsion. This repulsion causes the ionization energy to be sligh...
- Wed Oct 28, 2020 10:33 pm
- Forum: Photoelectric Effect
- Topic: UA Session Question
- Replies: 2
- Views: 181
Re: UA Session Question
Since the wavelength of the electrons is given (1.7 x 10^5 m), you can first find the velocity of the ejected electrons by using De Broglie's wave equation: \lambda = \frac{h}{mv} . After finding velocity of the ejected electron, you can find the kinetic energy of the ejected electron ( \frac{1}{2}m...
- Wed Oct 28, 2020 10:24 pm
- Forum: Einstein Equation
- Topic: Sapling HW #7
- Replies: 7
- Views: 460
Re: Sapling HW #7
From the photoelectric effect experiment, we know that E_{photon}- \Phi (work function) = E_{k} . Since you are calculating the maximum or longest wavelength possible that will eject electrons, you can assume that the kinetic energy is 0, so E_{photon} = \frac{hc}{\lambda } = \Phi , and you ...
- Wed Oct 28, 2020 6:54 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Atomic Spectra wave-like or particle-like properties
- Replies: 3
- Views: 93
Re: Atomic Spectra wave-like or particle-like properties
According to the Bohr frequency condition, when a high energy electron drops to a lower energy level, the difference in energy is given off as a photon, so light would be behaving like a particle in an experiment involving atomic spectra.
- Tue Oct 20, 2020 11:50 am
- Forum: Properties of Electrons
- Topic: Wave Properties of Electrons and Diffraction Patterns
- Replies: 2
- Views: 103
Re: Wave Properties of Electrons and Diffraction Patterns
Constructive interference (or waves in phase) means that the peak of one wave overlaps the peak of another wave, and the trough of one wave overlaps the trough of another wave. Destructive interference (waves out of phase) means that the peak of one wave overlaps with the trough of another. Waves in...
- Mon Oct 19, 2020 10:01 pm
- Forum: Student Social/Study Group
- Topic: Workshops
- Replies: 3
- Views: 150
Re: Workshops
Each workshop has problems separate from other workshops because each UA creates their own worksheets, so going to multiple workshops will give you more practice problems to do.
- Mon Oct 19, 2020 9:38 pm
- Forum: Balancing Chemical Reactions
- Topic: Formula Unit vs. Molecules
- Replies: 3
- Views: 1093
Re: Formula Unit vs. Molecules
I believe that formula units are used when referring to ionic compounds, and molecules are used when referring to molecular compounds. I hope this helps!
- Mon Oct 19, 2020 9:32 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Difference in Equations
- Replies: 4
- Views: 236
Re: Difference in Equations
En = -hr/(n^2) is an empirical equation for an H atom that can be used when finding the energy difference between two energy levels. For example, if you were asked to find the energy emitted when an electron makes a transition from the fourth to the second principle quantum level, you could use this...
- Mon Oct 19, 2020 9:21 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: Problem 1B.25
- Replies: 6
- Views: 215
Re: Problem 1B.25
For this problem, you would use Heisenberg's uncertainty equation: Δp x Δx ≥ h/(4π). We also know that Δp = m x Δv. Since the problem is asking for minimum uncertainty in the speed of an electron, we are trying to solve for Δv. The problem gives you Δx = 350. pm (but you should convert this to mete...
- Mon Oct 19, 2020 9:15 pm
- Forum: Photoelectric Effect
- Topic: Concept question photoelectric effect
- Replies: 1
- Views: 102
Re: Concept question photoelectric effect
The photoelectric effect tells us that light is made up of photons, and when trying to eject electrons from a metal surface, the energy of each photon must be sufficient to overcome the binding energy between the atom and the electron (or the threshold energy). Each photon interacts with one electro...
- Thu Oct 15, 2020 1:14 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Textbook Problem 1A.11
- Replies: 2
- Views: 92
Re: Textbook Problem 1A.11
Within a series, the final energy level is the same. For example, in the Balmer series, the lower or final energy level is n = 2, and in the Lyman series, the lower energy level is n = 1. Furthermore, the Balmer series is the visible light region, and the Lyman series is the ultraviolet region. I ho...
- Thu Oct 15, 2020 1:05 pm
- Forum: Einstein Equation
- Topic: Converting from MHz to Hz
- Replies: 5
- Views: 215
Re: Converting from MHz to Hz
1 MHz = 10^6 Hz, so I would check to make sure you converted MHz to Hz correctly. Also, I believe the answer for this problem should be 1 m. I hope this helps!