Search found 104 matches
- Mon Mar 15, 2021 11:47 am
- Forum: General Science Questions
- Topic: Final thoughts
- Replies: 28
- Views: 4480
Re: Final thoughts
I felt that the final was fair. There were a few questions that I found difficult, but it was balanced and the allotted time allowed for me to review all my answers multiple times. Looking forward to hopefully returning to in-person classes soon!
- Fri Mar 12, 2021 8:59 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Textbook 6L.7b [ENDORSED]
- Replies: 2
- Views: 295
Textbook 6L.7b [ENDORSED]
Write the half-reactions and devise a galvanic cell (write a cell diagram) to study each of the following reactions: b) H+(aq) + OH−(aq) → H2O(l),the Brønsted neutralization reaction The solution manual shows that the half reactions for this are O2(g) + 4H+(aq) + 4e- → 2H2O(l) O2(g) +2H2O(l) + 4e- →...
- Thu Mar 11, 2021 9:16 pm
- Forum: General Rate Laws
- Topic: Determining Intermediate Concentrations
- Replies: 6
- Views: 440
Re: Determining Intermediate Concentrations
An intermediate is a species that is produced, then consumed, so it would not be present in the overall rate law.
- Tue Mar 09, 2021 9:20 pm
- Forum: Experimental Details
- Topic: Reaction Profiles
- Replies: 2
- Views: 276
Re: Reaction Profiles
The second transition state is not necessarily always larger than the first one. The relative height is dependent on whether the elementary step it represents is slow or fast. For example, if the second step of a reaction is slow, then its transition state will be higher since it has a larger activa...
- Tue Mar 09, 2021 9:14 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Sapling #7
- Replies: 5
- Views: 362
Re: Sapling #7
When you type out the answer, it is not necessary to include the exponent for a term raised to the first power nor the [C] term since it will equal 1 given that it is raised to the zero power, so you rate law should be Rate = k[A][B]^2
- Tue Mar 09, 2021 8:32 pm
- Forum: Balancing Redox Reactions
- Topic: Textbook 6K.3
- Replies: 3
- Views: 235
Re: Textbook 6K.3
Hi! If you look in the solution manuals errors file, there is a typo in the question. The equation you are trying to balance should be Cl2(g) → ClOH(aq) + Cl-(aq). Hope this helps! https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14B/Solution_Manual_Errors_7Ed.pdf Ohh that makes a lot ...
- Tue Mar 09, 2021 5:52 pm
- Forum: Balancing Redox Reactions
- Topic: Textbook 6K.3
- Replies: 3
- Views: 235
Textbook 6K.3
I'm not quite sure how to balance the reaction in part d. The equation is Cl2(g) -> HClO(aq) + Cl2(g). I've tried adding water and hydrogen ions to each side but I can't seem to balance it. Is there any kind of method for balancing redox reactions with only one reactant? Thank you!
- Sun Mar 07, 2021 12:21 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Final on 3/13
- Replies: 18
- Views: 834
Re: Final on 3/13
The final is from 9:30-11am. You can find this on the exam schedule link on Dr. Lavelle's website!
- Sat Mar 06, 2021 5:42 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: q. 5 sapling
- Replies: 10
- Views: 574
Re: q. 5 sapling
You would take the same approach as you did for [A], since the concentration is tripled, you would put [3B].
- Sat Mar 06, 2021 1:40 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Molecularity
- Replies: 10
- Views: 423
Re: Molecularity
Yes! I think that's right, molecularity describes the number of molecules that participate in an elementary step, so the molecularity is equal to the sum of the stoichiometric coefficients of the reactants.
- Thu Mar 04, 2021 11:44 am
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: 6N.9 Question
- Replies: 2
- Views: 190
Re: 6N.9 Question
You need to look up the reduction potentials in Appendix 2B for H+ and Sn 2+. You can use this to solve for E naught and then plug this value and the concentrations given in the question into the Nernst equation to solve for -log[H+] which is equivalent to pH. Hopefully this helps you get started!
- Thu Mar 04, 2021 9:31 am
- Forum: Zero Order Reactions
- Topic: Sapling HW Week 9/10 #4
- Replies: 4
- Views: 357
Re: Sapling HW Week 9/10 #4
The unit for rate is molarity per second (M/s) so depending on the order of the reaction, you will determine the units for k that will cancel the other terms to give you the units for rate. For example, if you have a rate law: Rate = k[A]^2[B] the overall order of the reaction is 3, so you have M^3 ...
- Fri Feb 26, 2021 11:52 am
- Forum: Interesting Applications: Rechargeable Batteries (Cell Phones, Notebooks, Cars), Fuel Cells (Space Shuttle), Photovoltaic Cells (Solar Panels), Electrolysis, Rust
- Topic: Sapling Week 7/8 #18
- Replies: 7
- Views: 605
Re: Sapling Week 7/8 #18
Iron rusts in the presence of water and oxygen, so your unbalanced equation should be:
Fe + H2O + O2 -> Fe2O3∙3H2O
Now you can just add coefficients to balance the equation!
Fe + H2O + O2 -> Fe2O3∙3H2O
Now you can just add coefficients to balance the equation!
- Fri Feb 26, 2021 11:49 am
- Forum: Balancing Redox Reactions
- Topic: Sapling #5 Wk 7/8
- Replies: 3
- Views: 175
Re: Sapling #5 Wk 7/8
You haven't balanced the hydrogens! They are present on the left side of your equation, but not on the right.
- Thu Feb 25, 2021 4:52 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: When to use Different Nernst Equations
- Replies: 11
- Views: 832
Re: When to use Different Nernst Equations
I agree with the previous response that they are used interchangeably! I'd like to add that I believe when there is no longer a T variable, that the assumed temperature is 298K, so we would need to use E = E* - RT/nF(lnQ) if the reaction occurs at a different temperature.
- Thu Feb 25, 2021 4:47 pm
- Forum: Balancing Redox Reactions
- Topic: sapling q. 7
- Replies: 1
- Views: 115
Re: sapling q. 7
When using shorthand notation, solid species should be written on the outside and aqueous species should be towards the center closest to the salt bridge!
- Thu Feb 25, 2021 4:46 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Sapling Week 7&8 #7
- Replies: 1
- Views: 93
Re: Sapling Week 7&8 #7
Aqueous substances should be the closest to the salt bridge and the solids should be on the outside!
- Thu Feb 25, 2021 1:32 pm
- Forum: Balancing Redox Reactions
- Topic: Sapling #3 Week 7&8
- Replies: 2
- Views: 191
Re: Sapling #3 Week 7&8
Have you possibly typed out the equation wrong? NO-2 to NO-3 should be reduction since there is a gain of an electron. But here are some general steps for balancing redox reactions in basic solution (this method is different that the one given on Dr. Lavelle's website): 1) Balance the half reactions...
- Wed Feb 24, 2021 9:26 pm
- Forum: Balancing Redox Reactions
- Topic: Sapling Week 7/8 #18
- Replies: 1
- Views: 141
Sapling Week 7/8 #18
One of the most recognizable corrosion reactions is the rusting of iron. Rust is caused by iron reacting with oxygen gas in the presence of water to create an oxide layer. Iron can form several different oxides, each having its own unique color. Red rust is caused by the formation of iron(III) oxide...
- Sun Feb 21, 2021 9:53 pm
- Forum: Balancing Redox Reactions
- Topic: Sapling Week 7/8 #1
- Replies: 10
- Views: 2006
Re: Sapling Week 7/8 #1
Adding on to the previous reply, the oxidation number of the elements should add up to the overall charge of the molecule, so this will help you determine the oxidation number of an element after following the rules for cations, oxygen, etc.
- Tue Feb 16, 2021 9:23 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Textbook 4I.5
- Replies: 3
- Views: 245
Re: Textbook 4I.5
You just forgot to set the side of the equation with the 50°C water negative since it is losing heat. Specific heat capacity can be used!
- Tue Feb 16, 2021 9:20 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Which R value to choose
- Replies: 13
- Views: 1432
Re: Which R value to choose
I think the best way to determine which R value to use is to look at the units and find the one that will cancel to give you the units you need for your final answer. In the case for the ideal gas law, PV = nRT, you would choose 0.08206 L.atm.K^-1.mol^-1 since these are the units for pressure, volum...
- Tue Feb 16, 2021 9:16 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Textbook problem 4I.7
- Replies: 1
- Views: 118
Re: Textbook problem 4I.7
So for part A we know that ΔS(sys) = ΔHvap / (boiling point). Using the table, you can find these values for CH4 and plug them into the equation. ΔS(surr) will just be the negative value of ΔS(sys). Part B and C will follow the same process but with ΔHfus and freezing point. Keep in mind for part B ...
- Thu Feb 11, 2021 4:54 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Sapling Weeks 5 and 6 Question #14
- Replies: 3
- Views: 166
Re: Sapling Weeks 5 and 6 Question #14
ΔSvap = (ΔHvap) / (T @ BP) so you can just plug in the values given and solve for T!
- Thu Feb 11, 2021 4:52 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Sapling Weeks 5 and 6 Question #7
- Replies: 2
- Views: 168
Re: Sapling Weeks 5 and 6 Question #7
First you need to convert the watts to kJ/min which can be done through dimensional analysis. Then determine the moles of each sample that were vaporized by doing (initial mass) - (final mass) and dividing by the molar mass of the sample. Now that you know how long the sample was heated (3.12 minute...
- Wed Feb 10, 2021 10:09 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Textbook 4D.9
- Replies: 2
- Views: 243
Textbook 4D.9
The enthalpy of formation of trinitrotoluene (TNT) is −67kJ/mol, and the density of TNT is 1.65g/cm^3. In principle, it could be used as a rocket fuel, with the gases resulting from its decomposition streaming out of the rocket to give the required thrust. In practice, of course, it would be extreme...
- Wed Feb 10, 2021 9:51 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Exercise 4D Question 7
- Replies: 3
- Views: 90
Re: Exercise 4D Question 7
Yes! I think since they don't specify we can just assume 298K because that is the standard temperature.
- Tue Feb 09, 2021 5:29 pm
- Forum: Phase Changes & Related Calculations
- Topic: internal energy equation
- Replies: 5
- Views: 240
Re: internal energy equation
The correct equation is ΔU = q + w but there are different variations you can derive if there is constant pressure or constant volume. Possibly those are the equations you see online?
- Sun Feb 07, 2021 12:41 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Sapling Question #9 Week 3/4
- Replies: 3
- Views: 541
Re: Sapling Question #9 Week 3/4
There is heat transfer between the two volumes of water so the heat lost by the warmer water is equal to the heat gained by the cooler water: q(420.00mL water) = -q(130.00mL water) Use the formula, q = mc(Tf - Ti) and plug in the values given to solve for the final temperature! Hopefully this helps!
- Sat Feb 06, 2021 10:18 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Sapling Week 3/4 #19
- Replies: 6
- Views: 180
Re: Sapling Week 3/4 #19
Your first need to find the specific heat of the calorimeter which is done by dividing q cal by ΔT. The heat gained by the subsequent reaction is equal to the heat lost by the calorimeter so you can solve for q rxn by setting it equal to -q cal. Because this reaction occurs in constant volume ΔU = q...
- Wed Feb 03, 2021 9:54 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Week3&4 Sapling Q10
- Replies: 1
- Views: 90
Re: Week3&4 Sapling Q10
There is energy transfer between the ice and the water, so q(melt ice) + q(raise temp of melted ice) = -q(water). q(melt ice) = n(ΔHf water) Use q = mc(Tf - Ti) to find the energy required to change the temperature of the melted ice and water. Then just plug the values given in the problem to solve ...
- Wed Feb 03, 2021 8:34 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Week3&4 Sapling Q16
- Replies: 2
- Views: 163
Re: Week3&4 Sapling Q16
ΔE = w + q
w = 425 kcal x (4.184 kJ / kcal)
q = -5.00 x 10^2 kJ (this is negative because heat is being released)
ΔE = (425 kcal x (4.184 kJ / kcal)) - (5.00 x 10^2 kJ)
w = 425 kcal x (4.184 kJ / kcal)
q = -5.00 x 10^2 kJ (this is negative because heat is being released)
ΔE = (425 kcal x (4.184 kJ / kcal)) - (5.00 x 10^2 kJ)
- Tue Feb 02, 2021 9:32 pm
- Forum: Student Social/Study Group
- Topic: Test curve?
- Replies: 40
- Views: 2808
Re: Test curve?
Midterms were not curved in 14A because the test averages were fairly high, so I don't think they will be curved this quarter either.
- Sun Jan 31, 2021 6:46 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Sapling week 4 #12
- Replies: 1
- Views: 72
Re: Sapling week 4 #12
This question is asking you to find the heat capacity of the calorimeter, not the specific heat capacity of a substance. Heat capacity is the heat required to raise the temperature of an object by 1 degree. This stays constant for the calorimeter which is why you don't need to divide by grams.
- Sun Jan 31, 2021 3:27 pm
- Forum: Phase Changes & Related Calculations
- Topic: Sapling Week 3/4 Q10
- Replies: 3
- Views: 254
Re: Sapling Week 3/4 Q10
You can solve this by setting up an equation to represent the heat transfer between the ice cube and water: heat required to melt ice + heat required to change temp. of ice = heat released during temp. change in water. Use the heat of fusion of ice to determine the energy used to melt the ice. Use q...
- Sat Jan 30, 2021 3:12 pm
- Forum: Phase Changes & Related Calculations
- Topic: Sapling Week 3/4 Q4
- Replies: 10
- Views: 488
Re: Sapling Week 3/4 Q4
If the strength of the bonds on the reactants is larger, then the reaction is endothermic. If the strength of the bonds on the products is larger, then the reaction is exothermic. So, you can "add" the relative bond strengths on each side of the reaction to determine whether the reactants ...
- Wed Jan 27, 2021 4:26 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Textbook 5J5 (d)
- Replies: 2
- Views: 100
Re: Textbook 5J5 (d)
On Dr. Lavelle's website, he says that the correct equation is 2HD(g) ⇌ H2(g) + D2(g) so there is no change because there are equal moles of gas on each side.
- Tue Jan 26, 2021 8:15 pm
- Forum: Phase Changes & Related Calculations
- Topic: standard form
- Replies: 14
- Views: 695
Re: standard form
I think the ones we need to memorize are the diatomic molecules (H2, N2, O2, F2, Cl2, Br2, and I2), all of which are gases except for Br2 which is a liquid and I2 which is a solid. It would also be good to know that mercury is a liquid in its standard state and carbon is solid. If anyone knows any o...
- Sun Jan 24, 2021 10:17 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: hw question #2
- Replies: 10
- Views: 287
Re: hw question #2
You seem to be solving the problem correctly, I think your numbers are just off because you rounded the third term in the quadratic equation too soon. Try using more decimal places for that number and your answer should change.
- Sat Jan 23, 2021 11:41 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Averages of other molecules
- Replies: 3
- Views: 92
Re: Averages of other molecules
Bond enthalpies are only exact for bonds with the same atom (for example, a C-C bond), but for bonds between different atoms the bond enthalpy is an average of different bonds for different molecules. An example would be that a C-H bond would have slightly different bond enthalpies depending on the ...
- Thu Jan 21, 2021 9:22 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Including (aq) Solvents
- Replies: 4
- Views: 217
Re: Including (aq) Solvents
The solvent for aqueous solutions is most likely water, so it would not be included in the equilibrium constant. I think the best thing to do is just remember that liquids and solids are excluded from the expression.
- Wed Jan 20, 2021 8:07 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Midterm 1
- Replies: 9
- Views: 412
Re: Midterm 1
I believe everything that we have gone over up until this Friday's lecture will be tested on the midterm, so chemical equilibrium, acids/bases, and the beginning of thermochemistry. Lavelle will probably be sending out an email to confirm the topics too!Seraphina Joseph 1C wrote:What outlines are being tested on Midterm 1?
- Tue Jan 19, 2021 10:32 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Rounding E Values in ICE charts
- Replies: 17
- Views: 594
Re: Rounding E Values in ICE charts
Adding on to the previous reply, if you approximate by omitting x, you should also check that it is less than 5% of the initial concentration to make sure it is okay to approximate. Otherwise, the quadratic formula should be used to solve the equation.
- Fri Jan 15, 2021 7:43 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Percent ionization
- Replies: 6
- Views: 182
Re: Percent ionization
% ionization = [H3O+]eq / [HA]initial x 100%
- Fri Jan 15, 2021 7:41 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: percentage protonation for codeine
- Replies: 1
- Views: 63
Re: percentage protonation for codeine
When I solved this problem, I was also getting 1.5%. What I did was use the pKa to find Kb which I found to be 1.622x10^-6. Then I set this equal to (x^2)/(0.0073-x) and solved for x. I got x = 1.08x10^-4 = [OH-] (1.08x10^-4) / 0.0073 x 100% = 1.5% I'm not sure if I am doing the problem wrong or if ...
- Wed Jan 13, 2021 7:58 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Increase in Pressure
- Replies: 31
- Views: 699
Re: Increase in Pressure
This only applies to reactions with gases, so I don't believe there would be any problems involving a change in pressure with aqueous solutions.
- Wed Jan 13, 2021 3:33 pm
- Forum: Student Social/Study Group
- Topic: Study Tips for Chem 14B
- Replies: 11
- Views: 652
Re: Study Tips for Chem 14B
I also think that the textbook problems from the outline are very helpful and is great practice for exam questions. Another thing I would recommend is the UA sessions. They provide worksheets that have past exam problems and also are great at explaining the topics from lecture and answering any ques...
- Tue Jan 12, 2021 4:07 pm
- Forum: Administrative Questions and Class Announcements
- Topic: UA Workshops
- Replies: 6
- Views: 228
Re: UA Workshops
They are different for every UA!
- Sat Jan 09, 2021 5:44 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Sapling Week 1 #4
- Replies: 4
- Views: 151
Re: Sapling Week 1 #4
You need to make an ice table and use the initial pressure and K value to find all the partial pressures at equilibrium. PCl3 Cl2 PCl5 I 0 0 0.0230 C +x +x -x E x x 0.0230-x Once you have the partial pressures of each gas, add them together to find the total pressure.
- Fri Jan 08, 2021 9:45 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Chemical Equilibrium Part 2 Post Module #30
- Replies: 2
- Views: 103
Re: Chemical Equilibrium Part 2 Post Module #30
Yes, that's right, find the molarity of H2O by doing 2.5mol / 50L = 0.05M. You can disregard C because it is a solid and won't affect the equilibrium constant. Then you can put this value into the icebox and solve for the concentrations of CO and H2O at equilibrium and plug these values along with [...
- Tue Jan 05, 2021 10:00 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Question about 14B Exams
- Replies: 38
- Views: 1287
Re: Question about 14B Exams
I believe so! The UA said that everything should be the same as 14A with the exception of the change in timing for exams which would be taken in lecture rather than discussion.
- Tue Jan 05, 2021 9:58 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Chemical Equilibrium Part 2 Post-Assessment Problem
- Replies: 4
- Views: 109
Re: Chemical Equilibrium Part 2 Post-Assessment Problem
1. Balance the equation: 2AsH3 (g) <--> 2As (s) + 3H2 (g) 2. Solve for the concentration of AsH3 and H2 by dividing the given molar amounts by 3L (you can disregard As in this problem because solids are not included when solving for Q) 3. Plug in the values of the concentrations into the reaction qu...
- Tue Jan 05, 2021 10:39 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Audio Visual Assessment Question
- Replies: 3
- Views: 202
Re: Audio Visual Assessment Question
For this problem, you would use the ideal gas law, PV=nRT. Since you are given the concentrations of each gas, you can rearrange the equation by dividing both sides by V to get P=(n/V)RT. n/V is equal to concentration (mol/L) so plugging in the concentrations of the gases given into (n/V) of the equ...
- Thu Dec 10, 2020 11:01 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Bond angles
- Replies: 12
- Views: 792
Re: Bond angles
Yes! This is definitely something you should be familiar with.
- Thu Dec 10, 2020 10:07 pm
- Forum: Calculating the pH of Salt Solutions
- Topic: Why is Cu(NO3)2 Acidic?
- Replies: 2
- Views: 4812
Why is Cu(NO3)2 Acidic?
Question 6D.11 from the textbook asks: Decide whether an aqueous solution of each of the following salts has a pH equal to, greater than, or less than 7.
The answer key says that Cu(NO3)2 has a pH less than 7. How do we determine that?
The answer key says that Cu(NO3)2 has a pH less than 7. How do we determine that?
- Tue Dec 08, 2020 9:12 pm
- Forum: Calculating the pH of Salt Solutions
- Topic: Textbook 6B.5
- Replies: 4
- Views: 210
Re: Textbook 6B.5
I think if you are given a base, like in these questions, the best approach is to find the pOH first and use that to find the pH. You would first need to find the concentration of OH- from the given base and calculate the pOH using -log[OH-]. We know that pH + pOH = 14 so that's how you can solve fo...
- Tue Dec 08, 2020 9:04 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: 9C.3 Part C & D
- Replies: 1
- Views: 81
Re: 9C.3 Part C & D
They both represent water, so it doesn't really matter which way you decide to write it. The reason it is sometimes written as OH2 rather than the conventional way (H2O) is to show that it is the oxygen atom that attaches to the TM cation.
- Tue Dec 08, 2020 10:16 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: textbook #2E13
- Replies: 4
- Views: 300
Re: textbook #2E13
This molecule has trigonal bipyramidal electron geometry, so the three lone pairs would be trigonal planar while the I atoms would be axial forming a 180 degree bond angle.
- Fri Dec 04, 2020 8:08 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Geometry of Coordination Compounds
- Replies: 1
- Views: 89
Re: Geometry of Coordination Compounds
Yes! The geometry of a coordination compound is determined by the number of bonds on the TM cation, or the coordination number. I think there are many different geometries but we will mostly be dealing with either 4 or 6 bonds on the TM cation. If the coordination number is 6, the shape will be octa...
- Thu Dec 03, 2020 4:47 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Tetrahedral vs Square Planar
- Replies: 6
- Views: 301
Re: Tetrahedral vs Square Planar
Adding on, while we won't need to distinguish between square planar and tetrahedral for a coordination number of 4, I think there are a few specific coordination compounds we should know the shape of. I think it'll be helpful to memorize that cisplatin and the heme complex are square planar, but oth...
- Thu Dec 03, 2020 4:37 pm
- Forum: Naming
- Topic: Naming Order
- Replies: 16
- Views: 616
Re: Naming Order
I actually don't think the order for writing the chemical formulas matters according to what the UAs have said, but the convention seems to be writing the TM cation first and then the ligands next.
- Wed Dec 02, 2020 1:45 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Understanding coordination compounds
- Replies: 3
- Views: 175
Re: Understanding coordination compounds
I believe there has to 6 bonds for the shape to be octahedral. If there are only 4 bonds the shape would either be square planar or tetrahedral.
- Wed Dec 02, 2020 11:10 am
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: octahedral
- Replies: 4
- Views: 239
Re: octahedral
I'm not exactly sure what you mean, you may have heard incorrectly during lecture? During the lectures about coordination compounds, we learned that the number of bonds on a transition metal cation will determine its shape. The shape of the coordination compound will be octahedral if there are 6 bon...
- Fri Nov 27, 2020 2:35 pm
- Forum: Lewis Structures
- Topic: Question about AsO4 3- Lewis Structure
- Replies: 1
- Views: 130
Re: Question about AsO4 3- Lewis Structure
When drawing Lewis structures, we want to find the most stable molecule, and if the central atom can have an expanded octet, then that is a valid structure. Since As is in the third period, it can have and expanded octet, and by changing one of the As-O single bonds to a double bond, the molecule wi...
- Wed Nov 25, 2020 6:08 pm
- Forum: Hybridization
- Topic: Long Pairs/Double & Triple Bonds
- Replies: 9
- Views: 480
Re: Long Pairs/Double & Triple Bonds
Hybridization is determined by the number of regions of electron density, so lone pairs would be taken into account. Whether the bond between atoms is a single, double, or triple bond doesn't really matter because that bond will still be counted as one region of electron density. So, the number of e...
- Tue Nov 24, 2020 10:15 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Sapling Homework
- Replies: 2
- Views: 102
Re: Sapling Homework
Yes! The shape is cyclical when the carbon atoms form a ring. Then you would just attach the hydrogen atoms to the carbon atoms and choose the type of bonds accordingly to use the right number of electrons.
- Tue Nov 24, 2020 10:12 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: equatorial vs. axial atom positions
- Replies: 4
- Views: 156
Re: equatorial vs. axial atom positions
For the molecular shape trigonal bipyramidal, the atoms in the equatorial position will be the four atoms that are arranged horizontally and form trigonal planar geometry. The atoms in the axial position are the ones arranged vertically and form linear geometry with the center atom.
- Mon Nov 23, 2020 5:39 pm
- Forum: Sigma & Pi Bonds
- Topic: Single Bond Pi Bonds?
- Replies: 2
- Views: 109
Re: Single Bond Pi Bonds?
Single bonds can't be pi bonds because pi bonds are formed when two lobes of an orbital on one atom overlap with two lobes of an orbital of another atom. This forms a double bond where the atoms cannot rotate freely without breaking the bond.
- Sun Nov 22, 2020 11:57 am
- Forum: Sigma & Pi Bonds
- Topic: Sigma and Pi Bonds
- Replies: 24
- Views: 1079
Re: Sigma and Pi Bonds
I don't believe that is the case. A double bond consists of a pi bond but it is not called one. Can anyone confirm?
- Thu Nov 19, 2020 10:44 pm
- Forum: Dipole Moments
- Topic: 3F:5
- Replies: 3
- Views: 192
Re: 3F:5
I think you're right that this has to do with the fact that I has a much larger radius than F so it is more polarizable. Since it is more polarizable, the temporary dipoles will be larger, and CHI3 will have much stronger LDFs, causing the normal melting point to be higher.
- Wed Nov 18, 2020 10:41 pm
- Forum: Student Social/Study Group
- Topic: Lewis Structure
- Replies: 4
- Views: 202
Re: Lewis Structure
Elements located in the p-block of period 3 or later are capable of having expanded octets. Boron and aluminum are capable of having an incomplete octet (they are stable with 6 electrons).
- Wed Nov 18, 2020 10:36 pm
- Forum: Trends in The Periodic Table
- Topic: Trend for effective nuclear charge
- Replies: 3
- Views: 253
Re: Trend for effective nuclear charge
Effective nuclear charge increases across a period because the number of protons is increasing but the the number of shielding electrons stays the same. I'm not sure if there is a trend going down a group since the number of protons is increasing, but so is the number of shielding electrons.
- Mon Nov 16, 2020 6:15 pm
- Forum: Ionic & Covalent Bonds
- Topic: Textbook 2A. 9
- Replies: 2
- Views: 125
Re: Textbook 2A. 9
[Ar]3d6 shows the configuration of a neutral Cr atom, and to find the metal with a 2+ charge, you count forward two elements on the periodic table from Cr since a 2+ charge is obtained when the metal loses electrons. Fe is the element that would give you an identical electron configuration to Cr whe...
- Sat Nov 14, 2020 9:59 pm
- Forum: Resonance Structures
- Topic: Determining oxidation numbers?
- Replies: 9
- Views: 328
Re: Determining oxidation numbers?
This is a list of oxidation rules copied directly from Sapling, I found that it explained it all very clearly so hopefully this helps :) A neutral element that is not part of a compound has an oxidation state of zero. Monoatomic ions have oxidation states equal to their ionic charges. The sum of the...
- Thu Nov 12, 2020 11:56 am
- Forum: Formal Charge and Oxidation Numbers
- Topic: 2.A.13
- Replies: 5
- Views: 279
Re: 2.A.13
The electron will be removed from the outermost shell, so using the periodic table, you can see which orbital the valence electrons are located in and that's where the e- will be removed to form the +1 ions. Hopefully this helps!
- Thu Nov 12, 2020 11:39 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: Hydrogen bonding why it works?
- Replies: 4
- Views: 257
Re: Hydrogen bonding why it works?
Hydrogen bonding isn't actually a bond, it is a special case of dipole-dipole attraction between two molecules so the hydrogen atom will still only have one bond.
- Wed Nov 11, 2020 5:07 pm
- Forum: Formal Charge and Oxidation Numbers
- Topic: Sapling #9
- Replies: 5
- Views: 353
Re: Sapling #9
The sum of oxidation numbers for each element must add up to the overall charge of the molecule (so in this case, it should equal -1). Oxygen has an oxidation of -2, and since there are four oxygen atoms, the sum of oxidation states for oxygen is -8. Now, you can pick an oxidation state for the chlo...
- Mon Nov 09, 2020 8:17 pm
- Forum: Resonance Structures
- Topic: oxidation numbers in regards to resonance
- Replies: 2
- Views: 97
oxidation numbers in regards to resonance
I was doing the Sapling homework and on question 9, it asks which Lewis Structure is most plausible based on the oxidation number. I found the oxidation number but I'm not sure how to apply this to determine which resonance structure is the most likely. If anyone could help, thanks!
- Sun Nov 08, 2020 6:54 pm
- Forum: Ionic & Covalent Bonds
- Topic: Identifying Ionic and Covalent Bonds
- Replies: 10
- Views: 214
Re: Identifying Ionic and Covalent Bonds
Adding on to the previous reply, ionic bonds are between a metal and nonmetal (a transfer of electrons) while covalent bonds are between two nonmetals who share electrons.
- Sun Nov 08, 2020 2:48 pm
- Forum: Bond Lengths & Energies
- Topic: Sapling #4 Bond Length
- Replies: 2
- Views: 78
Sapling #4 Bond Length
What does it mean for a bond to have overwhelming or ample bond character when given the Lewis Structure and bond length data and how do we know which description fits the chemical bond given?
- Fri Nov 06, 2020 5:37 pm
- Forum: Ionic & Covalent Bonds
- Topic: Electron Affinity vs. Electronegativity
- Replies: 12
- Views: 1309
Re: Electron Affinity vs. Electronegativity
Electron affinity refers to the amount of energy released when an electron is added to a neutral atom in the gas phase, while electronegativity is how well an atom can attract electrons, so they are not the same thing.
- Fri Nov 06, 2020 5:35 pm
- Forum: Polarisability of Anions, The Polarizing Power of Cations
- Topic: Polarization
- Replies: 2
- Views: 79
Re: Polarization
Polarization is when there is a separation of the positive and negative charge. An ion has high polarizing power, correct me if I'm wrong, when it has a small radius and is highly charged. The nucleus will have a strong pull on the electrons, so it causes distortion.
- Tue Nov 03, 2020 9:00 pm
- Forum: Quantum Numbers and The H-Atom
- Topic: order of electron cnfiguration
- Replies: 13
- Views: 499
Re: order of electron cnfiguration
Adding on to the previous reply, when you look at the orbitals in the order of the periodic table, 4s comes before 3d so it will be occupied first, but once the 3d orbital is occupied, 4s will have a higher energy, so it can be written after 3s to demonstrate increasing energy levels.
- Sun Nov 01, 2020 11:47 pm
- Forum: Trends in The Periodic Table
- Topic: studying
- Replies: 12
- Views: 381
Re: studying
The UA sessions are super helpful and I also highly recommend the textbook problems! I also like to take notes on the textbook chapters in addition to the lecture just in case I missed anything and it also can be helpful to clear up anything I may be confused about.
- Sun Nov 01, 2020 3:11 pm
- Forum: Einstein Equation
- Topic: Textbook Question 1.3
- Replies: 2
- Views: 201
Re: Textbook Question 1.3
Oh I see, I think you may be right. I overcomplicated the problem and solved for frequency which has the unit s^-1 and multiplied that with the energy to get J.s^-1, but 750 watts should be correct from your explanation. Thank you!
- Sun Nov 01, 2020 2:43 pm
- Forum: Einstein Equation
- Topic: Textbook Question 1.3
- Replies: 2
- Views: 201
Textbook Question 1.3
The question asks: In each second, a certain lamp produces 2.4×10^21 photons with a wavelength of 633 nm. How much power (in watts) is produced as radiation at this wavelength (1W = 1J⋅s^−1)? However, the solution only provides the energy of the photons (750 J) which I was able to solve for, but I w...
- Sat Oct 31, 2020 10:47 pm
- Forum: Properties of Electrons
- Topic: Sapling #3
- Replies: 7
- Views: 320
Re: Sapling #3
Wavelength are set properties of a pulse of light, so it will not change when the number of photons changes. Each photon has the same wavelength and frequency. Energy, however, is proportional to the number of photons, so since there are 100 photons, the energy will be multiplied by 100.
- Fri Oct 30, 2020 9:11 pm
- Forum: Quantum Numbers and The H-Atom
- Topic: Calculating M(L)
- Replies: 2
- Views: 85
Re: Calculating M(L)
Yes! You have it right, M(L) ranges from L to negative L.
- Thu Oct 29, 2020 4:26 pm
- Forum: Einstein Equation
- Topic: energy of photon
- Replies: 10
- Views: 471
Re: energy of photon
Both are valid equations to find the energy of a photon. The one you use will depend on the information given during a specific problem. They can also be used together by doing hv = work function + KE
- Sun Oct 25, 2020 5:10 pm
- Forum: Accuracy, Precision, Mole, Other Definitions
- Topic: E.23 Part A and C
- Replies: 4
- Views: 285
Re: E.23 Part A and C
This isn't a mass percent problem, it only consists of dimensional analysis. So, for part A the process of solving the question would be: 3.00g CaBr2 x (1 mole CaBr2 / 199.197g) x (1 mole Ca2+ / 1 mole CaBr2) You would repeat the same steps for part C, just make sure to convert kg to g first. Hopefu...
- Thu Oct 22, 2020 8:40 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Sapling Question 16
- Replies: 3
- Views: 191
Re: Sapling Question 16
I believe this is the excited state of neon. You can tell because there are 10 electrons, but the 2p subshell has not been fill yet an electron is in the 3s orbital.
The ground state configuration is how you would normally write it for neon.
1s^2 2s^2 2p^6
Hopefully this helps!
The ground state configuration is how you would normally write it for neon.
1s^2 2s^2 2p^6
Hopefully this helps!
- Thu Oct 22, 2020 4:40 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: Conceptual difference between momentum and velocity
- Replies: 4
- Views: 464
Re: Conceptual difference between momentum and velocity
Velocity gives us the speed and direction of an object while momentum is the tendency of an object to stay in motion. Momentum is dependent on mass while velocity is independent of mass.
- Wed Oct 21, 2020 10:18 pm
- Forum: Properties of Light
- Topic: Wavelengths in Light Spectrum
- Replies: 9
- Views: 450
Re: Wavelengths in Light Spectrum
I am not 100% sure, but I think we need to memorize the order of the electromagnetic spectrum (radio waves to gamma rays) and to know that the visible light region has a wavelength ranging from 400nm to 700nm.
- Wed Oct 21, 2020 2:04 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Schrodinger's Wave Function
- Replies: 2
- Views: 123
Re: Schrodinger's Wave Function
Schrodinger's Wave Function tells us the probability of finding an electron in a certain position. I don't think we will be going into more detail about the equation other than the basic concepts discussed in today's lecture. For orbitals, I think we will be going more in depth in the next few lectu...
- Sat Oct 17, 2020 10:16 am
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Topic 1A #15
- Replies: 5
- Views: 195
Topic 1A #15
How would you go about solving this problem? I found the change in energy by using E = hc/lambda but I wasn't sure how to find the initial and final energy levels. In the ultraviolet spectrum of atomic hydrogen, a line is observed at 102.6 nm. Determine the values of n for the initial and final ener...
- Thu Oct 15, 2020 9:32 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Atomic Spectra Module #29
- Replies: 1
- Views: 55
Re: Atomic Spectra Module #29
First, to find the energy per photon, you would use the equation: E = hc/lambda = (6.63x10^-34 x 3.00x10^8) / 1850x10^-9 which gives you 1.075x10^-19 J per photon. To find the number of photons, you divide the total energy (given in the question) by the energy per photon. 11J / (1.075x10^-19 J/photo...
- Thu Oct 15, 2020 1:31 pm
- Forum: Properties of Light
- Topic: Light Intensity
- Replies: 8
- Views: 179
Re: Light Intensity
Yes, light intensity is the number of photons. Increasing intensity can cause more electrons to be ejected if the energy per photon is greater than or equal to the energy required to remove an electron.
- Tue Oct 13, 2020 8:53 pm
- Forum: Photoelectric Effect
- Topic: Photoelectric effect post-test #29
- Replies: 2
- Views: 95
Re: Photoelectric effect post-test #29
This question is asking for the energy needed to remove an electron from ONE sodium atom, so you must divide the work function by Avogadro's number.
(150.6 kJ/mol) / (6.022x10^23 atoms) x (1000 J/1 kJ)
= 2.501x10^-19 J
(150.6 kJ/mol) / (6.022x10^23 atoms) x (1000 J/1 kJ)
= 2.501x10^-19 J
- Tue Oct 13, 2020 8:33 pm
- Forum: Photoelectric Effect
- Topic: Photoelectric effect post-test #30
- Replies: 4
- Views: 159
Re: Photoelectric effect post-test #30
This question is asking us to find the frequency of the light so we can use the equation hv - (work function) = kinetic energy We already know the work function (2.501x10^-19 J) and the kinetic energy (1.99x10^-19 J) from the previous questions, so we can plug these values into the above equation an...
- Fri Oct 09, 2020 9:00 pm
- Forum: Empirical & Molecular Formulas
- Topic: Fundamentals L39.
- Replies: 4
- Views: 271
Fundamentals L39.
I've been having trouble getting the correct answer for this problem so if anyone could help thanks :) A 1.50-g sample of metallic tin was placed in a 26.45-g crucible and heated until all the tin had reacted with the oxygen in air to form an oxide. The crucible and product together were found to we...