Search found 102 matches

by Kaylee Messick 3J
Sat Mar 13, 2021 12:32 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: Nerst Equation
Replies: 3
Views: 40

Re: Nerst Equation

In the Nerst equation, the E naught value would be calculated from the list of standard potentials in Appendix 2B. This value can be found by writing out the half reactions, and identifying the anode and the cathode. By using these values, you can calculate the E naught of the cell, which can then b...
by Kaylee Messick 3J
Sat Mar 13, 2021 12:16 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: ∆G vs ∆G˚
Replies: 5
Views: 26

Re: ∆G vs ∆G˚

From the textbook, it says that when ∆G˚is negative, it means that K>1, which is stated instead of saying that it is spontaneous, which I believe may be what you are referring to. This is because ∆G, as you stated, would depend on the actual point in the reaction at the time of calculating the value...
by Kaylee Messick 3J
Fri Mar 12, 2021 9:21 pm
Forum: First Order Reactions
Topic: Form of a First Order Reaction
Replies: 3
Views: 29

Re: Form of a First Order Reaction

I'm not entirely sure about the answer to your question but from Lecture #24, it discussed how the rate law can be determined for the elementary steps from the molecularity, however it also noted that the overall rate law depended on the slowest step. I think that maybe this might apply to the quest...
by Kaylee Messick 3J
Fri Mar 12, 2021 8:42 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: Catalysts and intermediates
Replies: 10
Views: 49

Re: Catalysts and intermediates

A catalyst is something that speeds up the rate of the reaction (by lowering the activation energy), but it is not consumed in the reaction. This means it will appear again at the end of the reaction because it is not used. An intermediate is something that is created and used in the reaction, and i...
by Kaylee Messick 3J
Fri Mar 12, 2021 8:08 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Textbook 6L #3
Replies: 1
Views: 23

Re: Textbook 6L #3

Question 6L.3 asks to find the half reactions and to write the balanced equation. The half reactions can be found from the cell diagram, which has the anode on the left and the cathode on the right. Once you have the half reactions, you can use Appendix 2B to find the correct half reactions and the ...
by Kaylee Messick 3J
Sun Mar 07, 2021 9:06 pm
Forum: General Rate Laws
Topic: Why are there negative orders
Replies: 14
Views: 70

Re: Why are there negative orders

The rate law normally depends on the concentration of reactants, however, as stated above, the products can sometimes factor into this rate law due to equilibrium and the reverse reaction. In some cases, this would be reflected in the negative order in the rate law.
by Kaylee Messick 3J
Sun Mar 07, 2021 8:42 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Texbook 6M.13
Replies: 4
Views: 11

Re: Texbook 6M.13

For this problem, as stated above, you can use Appendix 2B in the textbook, which lists the Standard Potentials. You can first identify the half reactions for each equation, and which half reaction is being reduced and which is being oxidized. Then, you can use these half-reactions to identify the s...
by Kaylee Messick 3J
Sun Mar 07, 2021 8:30 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Textbook Question 6L.3
Replies: 2
Views: 25

Re: Textbook Question 6L.3

As stated above, for these problems you would use the fact that the cell diagram is similar to the reaction in that the anode is on the left and the cathode is on the right. This means that on the left side, the half reaction will show the electrons being released while on the right side, the half r...
by Kaylee Messick 3J
Sun Mar 07, 2021 7:58 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Textbook 6N.1
Replies: 3
Views: 9

Re: Textbook 6N.1

For this problem, you can first write out the half reactions. Then, using Appendix 2B, you can find the E°cell using the values from the anode and cathode. Then, this value can be substituted into the equation lnK = nFE°cell / (RT) along with the number of electrons (n, which is found from the half ...
by Kaylee Messick 3J
Sun Mar 07, 2021 7:47 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Concentration Cell
Replies: 3
Views: 32

Re: Concentration Cell

I also think this is because the concentration cell reaches equilibrium if the reactants are equal in concentration. This is why the porous disk is permeable to ions so that the reaction can continue occurring as the ions flow to the opposite side. To have a concentration cell, there needs to be cel...
by Kaylee Messick 3J
Sun Feb 28, 2021 7:32 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: Maximum Potential
Replies: 7
Views: 65

Maximum Potential

In Lecture #17, why is the electromotive force equal to the maximum potential difference?
by Kaylee Messick 3J
Sun Feb 28, 2021 7:09 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: X electrode and Y electrode
Replies: 4
Views: 17

Re: X electrode and Y electrode

I had the same reasoning in that I think because electrode Y was being reduced, it was gaining electrons which have a mass. For electrode X, it was being oxidized, and therefore was losing electrons. Hope that helps!
by Kaylee Messick 3J
Sun Feb 28, 2021 6:44 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: spontaneous reactions
Replies: 3
Views: 26

Re: spontaneous reactions

From the equation ΔG∘=−nFE∘cell, we can see that for the overall value to be negative, E∘cell must be positive. This relates back to what we learned with Gibbs Free Energy, which was when delta G was negative, the reaction would be spontaneous. In order for delta G to be negative, the standard cell ...
by Kaylee Messick 3J
Sun Feb 28, 2021 5:30 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: Sapling #12
Replies: 4
Views: 42

Re: Sapling #12

This question is using the equation ΔG∘=−nFE∘cell. The number of moles is the moles of electrons that are transferred. To find E∘cell, you can use the table of Standard Reduction Potentials, which gives you the overall cell potential for the reaction. You can substitute these values into the equatio...
by Kaylee Messick 3J
Sun Feb 28, 2021 5:19 pm
Forum: Balancing Redox Reactions
Topic: Sapling #10
Replies: 6
Views: 23

Re: Sapling #10

For this question, you would have to use the Standard Reduction Potentials given in the table. Like stated above, the more negative values indicate a poorer oxidizing agent while the more positive values indicate a stronger oxidizing agent. Hope this helps!
by Kaylee Messick 3J
Sun Feb 21, 2021 3:08 pm
Forum: Balancing Redox Reactions
Topic: Textbook 6K 1 part c
Replies: 3
Views: 27

Re: Textbook 6K 1 part c

As stated above, in an acidic solution, you use H2O to balance the oxygen, and you use H+ to balance the hydrogen. After balancing the hydrogen and oxygen, you also have to balance the charge on each side, taking into account the coefficients on each component and their overall charge. You can add e...
by Kaylee Messick 3J
Sun Feb 21, 2021 2:48 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 4D. 9
Replies: 2
Views: 28

Re: 4D. 9

As stated above, you can find the standard enthalpies of formation values from Appendix 2. This value represents the total enthalpy for the reaction stated in the question, however, this is for 4 moles of TNT. To calculate for one mole of TNT, you would divide the total enthalpy by 4. Using this val...
by Kaylee Messick 3J
Sun Feb 21, 2021 2:38 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Salt Bridge Purpose
Replies: 8
Views: 45

Re: Salt Bridge Purpose

From lecture, the purpose of the salt bridge was to keep both of the solutions that the anode and cathode are in neutral. This is because as the electrons move from the left to the right side, the oxidized side becomes more positive and the reduced side becomes more negative. This means the electron...
by Kaylee Messick 3J
Sun Feb 21, 2021 11:10 am
Forum: Balancing Redox Reactions
Topic: redox example from lec 17
Replies: 5
Views: 42

Re: redox example from lec 17

The half reactions are basically a way to represent the gain or loss of electrons. In this case, the reason why we would write +5e would be because the 5Fe2+ is separating into 5Fe3+ and 5e-. If the products on the right side were added together, they would equal 5Fe2+. You can think of the Fe2+ as ...
by Kaylee Messick 3J
Sun Feb 21, 2021 10:57 am
Forum: Balancing Redox Reactions
Topic: Acidic and Basic Solutions
Replies: 1
Views: 33

Acidic and Basic Solutions

The textbook lists out the steps of balancing redox reactions in Toolbox K.1. In step 4, I understand that the reaction uses H2O and H+ when balancing a reaction in acidic solutions, and uses H2O and OH- when balancing in basic solutions. However, after that it states, "When ...OH− ...→ ... H2O...
by Kaylee Messick 3J
Sun Feb 14, 2021 9:12 pm
Forum: Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy
Topic: Statistical Entropy vs Residual Entropy
Replies: 1
Views: 12

Statistical Entropy vs Residual Entropy

I was confused on the difference between statistical entropy (which uses Boltzmann's formula), and residual entropy? They both seem to depend on the orientation of the molecule, but I'm not sure what other differences apply. Thanks!
by Kaylee Messick 3J
Sun Feb 14, 2021 9:00 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Sapling #6
Replies: 6
Views: 32

Re: Sapling #6

This problem can be calculated by solving for entropy with the change in volume and the change in temperature separately and adding them together. This is because entropy is a state function and different values can be added together to get the final answer. As stated above, you can use the equation...
by Kaylee Messick 3J
Sun Feb 14, 2021 8:25 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Sapling Week 5/6 #15
Replies: 5
Views: 28

Sapling Week 5/6 #15

Given the information A+B⟶2D C⟶D ΔH∘=737.6 kJ ΔS∘=335.0 J/K ΔH∘=526.0 kJ ΔS∘=−232.0 J/K calculate ΔG∘ at 298 K for the reaction A+B⟶2C I'm having trouble calculating this question. Does it matter if you calculate each delta H and delta S first then put the final values in the Gibbs Free Energy equat...
by Kaylee Messick 3J
Sun Feb 14, 2021 7:58 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Gibbs Free Energy and phase changes
Replies: 2
Views: 34

Re: Gibbs Free Energy and phase changes

To add on to what was mentioned above, because the molar entropy of a gas phase is greater than the liquid phase, when subtracting this value from delta H (from the Gibbs Free Energy equation), the value of delta G decreases more for the larger value of T and entropy. This would probably be why ther...
by Kaylee Messick 3J
Sun Feb 14, 2021 4:34 pm
Forum: Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy
Topic: Entropy trends from 4H
Replies: 8
Views: 69

Re: Entropy trends from 4H

Just to add on to what others have said, larger molecules have a higher mass and are more complex, and therefore have higher entropy. The textbook also references the different phases as an example. Gases have higher molar entropy because they can move around more (less rigid) while solids have lowe...
by Kaylee Messick 3J
Sun Feb 07, 2021 3:49 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Sapling #19 - Week 3/4
Replies: 3
Views: 32

Sapling #19 - Week 3/4

The problem states: "A constant‑volume calorimeter was calibrated by carrying out a reaction known to release 0.626 kJ of heat in 0.500 L of solution in the calorimeter (q =−0.626 kJ), resulting in a temperature rise of 2.00 ∘C. In a subsequent experiment, 250.0 mL of 0.20 M HClO2(aq) and 250.0...
by Kaylee Messick 3J
Sun Feb 07, 2021 3:27 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Sapling week 3/4 Question 5
Replies: 9
Views: 51

Re: Sapling week 3/4 Question 5

To answer this question, it is easiest to look for the equations that have the reactants and products (in the correct states) in them. Then, you can figure out what needs to be done to each enthalpy value of the corresponding equation, such as multiplying, reversing, or simply adding the delta H, in...
by Kaylee Messick 3J
Sun Feb 07, 2021 10:26 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Sapling #8 Week3/4
Replies: 6
Views: 50

Re: Sapling #8 Week3/4

For this question, I divided the heat absorbed by the amount of heat absorbed for one mole of CS2. This will provide the number of moles of CS2 produced, and then you can solve for grams. Hope this helps!
by Kaylee Messick 3J
Sun Feb 07, 2021 10:06 am
Forum: Calculating Work of Expansion
Topic: Sapling HW Week 3/4 #14 Part A
Replies: 2
Views: 35

Re: Sapling HW Week 3/4 #14 Part A

For part A, I think we only use the first given volume to calculate the moles of gas (not delta V, as in your first equation), so this would be 2.65 L in your case. After that, you are correct in that you substitute it in the equation for work. Hope this helps!
by Kaylee Messick 3J
Sun Feb 07, 2021 9:32 am
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: delta U vs delta H [ENDORSED]
Replies: 1
Views: 31

delta U vs delta H [ENDORSED]

What is the difference between delta U and delta H of a system, and when would they be equal?
by Kaylee Messick 3J
Sun Jan 31, 2021 7:36 pm
Forum: Calculating Work of Expansion
Topic: Sapling Week 3 and 4 #13
Replies: 2
Views: 28

Sapling Week 3 and 4 #13

The question states, "At constant pressure, which of these systems do work on the surroundings?" One of the correct answers is A(s) + B(g) ->2C(g). Does this mean that we only consider moles of gas when figuring out the work in a system?
by Kaylee Messick 3J
Sun Jan 31, 2021 7:12 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Isothermal, Reversible, and Irreversible Processes
Replies: 2
Views: 29

Re: Isothermal, Reversible, and Irreversible Processes

From the textbook, one of the reversible changes is the reversible isothermal expansion, in which isothermal is defined as constant-temperature. It states that the system is in a constant-temperature water bath during the expansion to stay at this temperature. Hope this helps!
by Kaylee Messick 3J
Sun Jan 31, 2021 7:02 pm
Forum: Calculating Work of Expansion
Topic: Variables in Reversible Expansion
Replies: 1
Views: 18

Variables in Reversible Expansion

The textbook defines reversible expansion as "infinitely small changes in a variable". I was wondering which variables in the formula for reversible expansion would be changing at a given time and which ones would be considered constant? Thanks!
by Kaylee Messick 3J
Thu Jan 28, 2021 8:12 pm
Forum: Acidity & Basicity Constants and The Conjugate Seesaw
Topic: Textbook Problem 6.19
Replies: 1
Views: 22

Textbook Problem 6.19

6.19) Hemoglobin (Hb) molecules in blood carry O2 molecules from the lungs, where the concentration of oxygen is high, to the tissues where it is low. In the tissues, the equilibrium H3O+ (aq) +HbO2- (aq) --> HHb (aq) + H2O (l) + O2 (aq) releases oxygen. When muscles work hard, they release lactic a...
by Kaylee Messick 3J
Thu Jan 28, 2021 7:37 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Textbook 4D.21, part c
Replies: 2
Views: 16

Re: Textbook 4D.21, part c

Hello! I believe that this is one of the problems with a typo in the solution. The correct answer should be on the document titled "Solution Manual Errors 7th Edition" on his website. Hope this helps!
by Kaylee Messick 3J
Sun Jan 24, 2021 8:29 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Making the Approximating for X: K or % Ionization?
Replies: 6
Views: 51

Re: Making the Approximating for X: K or % Ionization?

I would think that the percent ionization is most likely more helpful as an indicator because as seen in the Sapling question, sometimes the approximation using the value of K doesn't always work, so it is safer to check to make sure it is less than 5%. Hope this helps!
by Kaylee Messick 3J
Sun Jan 24, 2021 7:30 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Sapling Week 2 #5
Replies: 4
Views: 60

Re: Sapling Week 2 #5

I think that you had to add the two values because we were finding the concentration of amine after the reaction had already occurred (the products of [BH+] and [OH-] had formed already). This means that to find the original concentration of B, we had to add back the amount that was already protonat...
by Kaylee Messick 3J
Sun Jan 24, 2021 6:47 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Factors that affect the Equilibrium Constant
Replies: 8
Views: 49

Re: Factors that affect the Equilibrium Constant

The equilibrium constant is not affected by a change in concentration or a change in pressure. This means that a change in either the concentration or pressure would cause a change in the reactants or products, depending on what change occurred, in order to maintain the fixed ratio of the equilibriu...
by Kaylee Messick 3J
Sun Jan 24, 2021 6:14 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Sapling Question 2
Replies: 6
Views: 67

Re: Sapling Question 2

To find the percent ionization, you can use an ICE table to find the concentration of the products from the reaction using the Ka value provided. Then this value can be divided by the original amount provided and multiplied by 100 to find the percent ionization. These are similar to the problems in ...
by Kaylee Messick 3J
Sun Jan 24, 2021 9:40 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Sapling Week 2 Question 3
Replies: 6
Views: 51

Sapling Week 2 Question 3

For this question, the Ka = 8.40x10^-4. When solving this problem, I decided to use approximation, even though the Ka was at 10^-4. I got a percent ionization that was larger than 5%, so I went back to solve the problem with the quadratic formula. I got essentially the same answer as my approximatio...
by Kaylee Messick 3J
Sun Jan 17, 2021 6:51 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Sapling #4
Replies: 6
Views: 46

Re: Sapling #4

For this problem, I set up an ICE box and used it to find x. The initial pressure of PCl5 would be 0.0370 bar and the change would be -x, while the other initial pressures would be 0, and the change would be +x. After using Kp and finding x, you would calculate the pressures at equilibrium and add t...
by Kaylee Messick 3J
Sun Jan 17, 2021 6:27 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Neutral Solution
Replies: 1
Views: 13

Neutral Solution

In the lecture on Friday, it was mentioned that if [H3O+] < 10^-7, then the solution would be considered neutral. I am still a little confused as to why this would be and how we would apply this when solving a problem. Thank you!
by Kaylee Messick 3J
Sun Jan 17, 2021 3:07 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: shifts left or right
Replies: 23
Views: 110

Re: shifts left or right

The shift would indicate which way the reaction is going to go in order to be at equilibrium again. More reactants or products will form depending on the value of Q, which is important because the value of Q can be at anytime during the reaction, so comparing this value to K will indicate which dire...
by Kaylee Messick 3J
Sun Jan 17, 2021 10:25 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Equilibrium constant
Replies: 4
Views: 14

Re: Equilibrium constant

To answer the second question, a change in the concentrations of products and reactants do not change the equilibrium constant because the reaction will "adjust" itself to account for the changes in concentrations. This means that the amounts of reactants and products may change but the ov...
by Kaylee Messick 3J
Sun Jan 17, 2021 10:16 am
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Autoprotolysis Reaction and Kw
Replies: 3
Views: 12

Autoprotolysis Reaction and Kw

In the textbook in Section 6A.4, there is a Thinking Point question that asks, "The autoprotolysis reaction is endothermic; do you expect Kw to increase or decrease with increasing temperature?" I was wondering how temperature would affect Kw, because I thought that Kw was a constant. Than...
by Kaylee Messick 3J
Sun Jan 10, 2021 7:17 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Sapling number 6 Week 1
Replies: 3
Views: 33

Re: Sapling number 6 Week 1

To set up this problem, you would start with the equation for finding the reaction quotient, which is essentially the same as finding the equilibrium constant, K. For this problem, it would be [NH3]^2/[N2][H2]^3. Then, you would substitute the given molar concentrations and solve for Q. Hope this he...
by Kaylee Messick 3J
Sun Jan 10, 2021 6:44 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Equilibrium Adjustments
Replies: 10
Views: 93

Re: Equilibrium Adjustments

I think that this basically means that if there is a change to the reaction such as in pressure or concentration of reactants and products, the reaction will shift a certain direction to favor either the reactants or products to reach equilibrium again. Hope this helps!
by Kaylee Messick 3J
Sun Jan 10, 2021 6:17 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Stable Reactants and Products
Replies: 7
Views: 43

Re: Stable Reactants and Products

I agree with what the others have said! It is more likely that a product or reactant would form if it was more stable. This means that calculating the equilibrium constant and looking at its size can tell us which side (the reactants or products) is more stable, and therefore, favored.
by Kaylee Messick 3J
Sun Jan 10, 2021 5:30 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Question on #19 from focus topic 3
Replies: 3
Views: 21

Re: Question on #19 from focus topic 3

For this problem, make sure the reaction is balanced before starting the calculations. Then, convert the moles of reactant and product of the gases (the solid would not be included in finding the value of the reaction quotient) into the molar concentrations (moles divided by liters). Then, you can s...
by Kaylee Messick 3J
Sun Jan 10, 2021 2:52 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Homogeneous and Heterogeneous Equilibrium
Replies: 13
Views: 160

Homogeneous and Heterogeneous Equilibrium

Do the terms homogeneous and heterogeneous equilibrium only apply to the phases of the reactants and products used in the equation for K or are other reactants and products (such as solids and liquids) in the overall reaction taken into consideration when determining if the reaction is homogeneous o...
by Kaylee Messick 3J
Sat Dec 12, 2020 10:31 pm
Forum: Hybridization
Topic: exceptions ?
Replies: 3
Views: 70

Re: exceptions ?

Like the others said, the regions of electron density show the number of hybrid orbitals. This would mean that double bonds would still remain one region of electron density. For example, in the lecture on Friday, the example of CO2 had a hybridization of sp even though carbon had double bonds. Hope...
by Kaylee Messick 3J
Sat Dec 12, 2020 10:20 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: Sapling 10
Replies: 6
Views: 64

Re: Sapling 10

It also helps to look at the electronegativity values for the atoms or the number of atoms because they may end up pulling electrons away and influencing the strength of the acid/base.
by Kaylee Messick 3J
Sat Dec 12, 2020 4:30 pm
Forum: Polyprotic Acids & Bases
Topic: Amphiprotic
Replies: 2
Views: 33

Re: Amphiprotic

Amphiprotic means that molecule can both accept and donate protons. This is different from amphoteric which means that molecule can act as both an acid and a base, but might not necessary involve protons. In the textbook, they gave the example of water as an amphiprotic molecule. Hope this helps!
by Kaylee Messick 3J
Sat Dec 12, 2020 1:40 pm
Forum: Amphoteric Compounds
Topic: Textbook Question 6A.17
Replies: 2
Views: 29

Re: Textbook Question 6A.17

In the textbook, in Section 6A.3, it discusses oxides. It states that usually if there is a metal, it will be basic and if there is a nonmetal, it will be acidic. It also lists examples of amphoteric compounds. Hope this helps!
by Kaylee Messick 3J
Sat Dec 12, 2020 10:58 am
Forum: Bronsted Acids & Bases
Topic: lewis vs bronsted
Replies: 8
Views: 84

Re: lewis vs bronsted

A Lewis acid will accept an electron pair while a Lewis base will donate an electron pair. A Bronsted acid will donate a proton while a Bronsted base will accept a proton. I think of it as the Lewis acid will be accepting electrons (making it more negative) and a Bronsted acid will be donating a pro...
by Kaylee Messick 3J
Sun Dec 06, 2020 9:30 pm
Forum: Shape, Structure, Coordination Number, Ligands
Topic: Detrmining Shape from coordination numbers
Replies: 6
Views: 63

Re: Detrmining Shape from coordination numbers

Section 9C.2 from the textbook described the shapes certain complexes have and provides examples which helped me on this topic. Hope this helped!
by Kaylee Messick 3J
Sun Dec 06, 2020 8:51 pm
Forum: Lewis Acids & Bases
Topic: CH4 and NH3
Replies: 3
Views: 19

Re: CH4 and NH3

I'm not sure about CH4 but for NH3, it says in the textbook that it is a Bronsted base because when mixed with water, it ends up accepting a proton, meaning it is a proton acceptor, which is a base. Hope this helps!
by Kaylee Messick 3J
Sun Dec 06, 2020 8:22 pm
Forum: Naming
Topic: Naming Rundown
Replies: 2
Views: 29

Re: Naming Rundown

The textbook also states that if the ligand is an anion, then if they end in -ide, -ate, or -ite, it would be changed to -ido, -ato, or -ito. Hope this helps!
by Kaylee Messick 3J
Sun Dec 06, 2020 8:02 pm
Forum: Formal Charge and Oxidation Numbers
Topic: Determining Oxidation Numbers
Replies: 6
Views: 62

Re: Determining Oxidation Numbers

I remember it as the charge of the atom/molecule that will make the charge for the entire compound equal to what was stated in the problem. So I would add all the known charges and find the difference between that number and the entire charge of the compound. Hope this helps!
by Kaylee Messick 3J
Sun Dec 06, 2020 7:09 pm
Forum: Formal Charge and Oxidation Numbers
Topic: Sapling #6
Replies: 3
Views: 30

Re: Sapling #6

The way I think of it is the charge of the compound inside the brackets has to equal the opposite of the atom outside the brackets. Since the chlorine atom has a charge of -1, the charge of the atoms inside the brackets must be +1. This means since the chlorine inside the brackets is -1, the cobalt ...
by Kaylee Messick 3J
Sun Nov 29, 2020 9:56 pm
Forum: Hybridization
Topic: Hybridization
Replies: 3
Views: 23

Re: Hybridization

Hybridization allows for the representation of the correct shape. For example, when looking at the electron configuration of carbon, it looks as if there can only be two bonds formed, however in reality, carbon can form four bonds. Instead of having separate s and p orbitals, they can combine to for...
by Kaylee Messick 3J
Sun Nov 29, 2020 9:47 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Polar/Nonpolar Bonds vs Molecules
Replies: 4
Views: 46

Re: Polar/Nonpolar Bonds vs Molecules

For me, it helps to draw out the Lewis structures to determine the shape of the molecule. It also helps to identify any lone pairs because that will affect the bond angles, which will affect the shape. Hope this helps!
by Kaylee Messick 3J
Sun Nov 29, 2020 9:19 pm
Forum: Hybridization
Topic: How to find the hybridization
Replies: 14
Views: 116

Re: How to find the hybridization

I agree with what was said above. To add on, in Lecture #23, there is a list of the different orbitals and how they correlate to the hybridized orbitals and the shape of the molecule. Hope this helps!
by Kaylee Messick 3J
Sun Nov 29, 2020 9:02 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Polar
Replies: 10
Views: 100

Re: Polar

Whether a molecule is polar or nonpolar depends on if the dipoles cancel. The lone pairs can affect whether it is polar because having lone pairs affects the shape of the molecule and might mean that the atoms are not longer directly across from one another, meaning the dipoles do not cancel.
by Kaylee Messick 3J
Sun Nov 29, 2020 6:49 pm
Forum: Hybridization
Topic: Hybridization of Carbon
Replies: 10
Views: 106

Re: Hybridization of Carbon

I believe the number 3 comes from when one of the electrons in the s-orbital moves the unfilled p-orbital, so that there are 3 unpaired electrons. This means that carbon is able to form 4 bonds with the new sp3 hybrid orbitals. Hope this helps!
by Kaylee Messick 3J
Sun Nov 22, 2020 9:40 pm
Forum: Sigma & Pi Bonds
Topic: Sigma Bond Flexibility
Replies: 5
Views: 36

Re: Sigma Bond Flexibility

I agree with what was mentioned above. Only one sigma bond is formed because it forms from atoms end to end. This differs from pi bonds, which form in atoms that are side by side. When more than one bond forms, it is harder for the bonds to rotate because if they rotate against each other they will ...
by Kaylee Messick 3J
Sun Nov 22, 2020 9:22 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: memorizing VSEPR models
Replies: 13
Views: 111

Re: memorizing VSEPR models

In the textbook, in section 2E.1, there are images of all of the shapes next to each other which helps when differentiating between shapes. Hope this helps!
by Kaylee Messick 3J
Sun Nov 22, 2020 9:09 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Lone Pairs and Bonding Pairs
Replies: 6
Views: 41

Lone Pairs and Bonding Pairs

Sapling Question 6 asks which molecules are linear. One of the molecules I chose was SF2, which turned out to be incorrect. I reasoned that even though SF2 has two lone pairs, the name of the molecule is based only on the bonding pairs, which would mean that the two bonding pairs in this molecule ma...
by Kaylee Messick 3J
Sun Nov 22, 2020 8:45 pm
Forum: Hybridization
Topic: sigma and pi bonds in triple bonds
Replies: 2
Views: 22

Re: sigma and pi bonds in triple bonds

I think also that sigma bonds form between one end of one atom to the other end of a different atom, meaning there can only be one sigma bond in general. The pi bonds form between atoms that are side by side, which means there can be more of them. Hope this helps!
by Kaylee Messick 3J
Sun Nov 22, 2020 5:52 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Trigonal Planar vs Angular
Replies: 1
Views: 22

Trigonal Planar vs Angular

In the textbook in topic 2E, it was describing the shape of NO2, which has an unpaired electron. It describes the electron arrangement as trigonal planar, but atom arrangement as angular. What would be the difference between these two shapes, and which is the correct one to use? Thank you!
by Kaylee Messick 3J
Sun Nov 15, 2020 8:40 pm
Forum: Ionic & Covalent Bonds
Topic: Sapling week 5 & 6 #16
Replies: 7
Views: 51

Re: Sapling week 5 & 6 #16

Carbon and hydrogen atoms have similar electronegativity values and therefore would share electrons equally. This means they would not form hydrogen bonds because the atoms do not have partial charges. This eliminates B,C,E, and F. Answer choice A does not have a hydrogen atom so it would not form a...
by Kaylee Messick 3J
Sun Nov 15, 2020 7:50 pm
Forum: Ionic & Covalent Bonds
Topic: Dipole- Induced Dipole vs Induced Dipole - Induced Dipole
Replies: 1
Views: 14

Re: Dipole- Induced Dipole vs Induced Dipole - Induced Dipole

I believe that dipole-induced dipole is when one molecule has regions of positive and negative charges (meaning electrons are distributed unevenly), and when they are near a molecule that shares electrons equally, they can cause the electrons in that molecule to shift toward the molecule with the ch...
by Kaylee Messick 3J
Sun Nov 15, 2020 5:24 pm
Forum: Lewis Structures
Topic: Sapling #18
Replies: 2
Views: 53

Re: Sapling #18

In one of the lectures when Dr. Lavelle discussed how the shape of molecules affects the interactions between them, he drew two molecules similar to the ones in #18. Based off of this example, I drew CH3C(CH3)2CH3 with one carbon in the middle and the rest of the carbon molecules around it. Then I d...
by Kaylee Messick 3J
Sun Nov 15, 2020 4:40 pm
Forum: Ionic & Covalent Bonds
Topic: What do delta positive and delta negative refer to?
Replies: 3
Views: 37

Re: What do delta positive and delta negative refer to?

To answer your question about ionic and covalent bonding, in an ionic bond, there would be a gain or loss of electrons, making the entire molecule have a charge. In a covalent bond, which is when the electrons are being shared, there might be equal or unequal sharing of electrons. When electrons are...
by Kaylee Messick 3J
Sun Nov 15, 2020 4:07 pm
Forum: Formal Charge and Oxidation Numbers
Topic: Sapling week 5 & 6 #5
Replies: 20
Views: 148

Re: Sapling week 5 & 6 #5

I got that the charge on carbon is -2 because the overall molecule must not have a charge. So because there are two positive charges, carbon would be -2 to offset that. Hope this helps!
by Kaylee Messick 3J
Sun Nov 08, 2020 10:32 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Textbook 1E.13
Replies: 3
Views: 25

Re: Textbook 1E.13

I think silver is an exception, similar to how copper, which is directly above it, is an exception. The atom would rather have a half full (d^5) or full (d^10) subshell than (d^4) or (d^9). I'm not entirely sure the complete explanation, but I think it is because when the subshell is half full or fu...
by Kaylee Messick 3J
Sun Nov 08, 2020 10:24 pm
Forum: Ionic & Covalent Bonds
Topic: Sapling Question #3
Replies: 6
Views: 67

Re: Sapling Question #3

In one of the Workshops, they said that P, S, and Cl atoms often are exceptions to the octet rule, which helps when trying to draw Lewis structures and calculate formal charge.
by Kaylee Messick 3J
Sun Nov 08, 2020 9:11 pm
Forum: Ionic & Covalent Bonds
Topic: Sapling Arrange Bonds Ionic to Covalent
Replies: 3
Views: 50

Re: Sapling Arrange Bonds Ionic to Covalent

Dr. Lavelle stated in his last lecture that a guideline to follow to know if it is an ionic or covalent bond is if the electronegativity difference is greater than 2, then the bond is most likely ionic. If the electronegativity difference is less than 1.5, then it is most likely covalent. To look at...
by Kaylee Messick 3J
Sun Nov 08, 2020 8:55 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: question 2 sapling homework
Replies: 2
Views: 11

Re: question 2 sapling homework

I'm not sure if your second question specifically asks to put the structure in brackets, but the way to add formal charges is to use the " +" and " - " buttons in the More tab when drawing the structure. Then you just click on the atom to add the charge. Hope this helps!
by Kaylee Messick 3J
Sun Nov 08, 2020 8:49 pm
Forum: Resonance Structures
Topic: Formal charge on different resonance structures
Replies: 3
Views: 18

Re: Formal charge on different resonance structures

To add on, the structure with the least amount of formal charge is the most stable if all of the structures happen to have different formal charges. You would also want to make sure that the negative formal charge is on the most electronegative atom because these atoms can "hold on" to the...
by Kaylee Messick 3J
Sun Nov 01, 2020 7:30 pm
Forum: SI Units, Unit Conversions
Topic: Sapling WEEK 4 #22
Replies: 5
Views: 76

Re: Sapling WEEK 4 #22

Hello! For this problem, you would use the wavelength provided to find the velocity using the de Broglie equation. Then you would use this velocity to find the kinetic energy in joules, and convert this answer to eV. Hope this helps!
by Kaylee Messick 3J
Sun Nov 01, 2020 6:49 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Textbook Exercise 1E.15
Replies: 6
Views: 48

Re: Textbook Exercise 1E.15

Hello! I believe it would be vanadium, though I could be wrong because I am still a little confused on what differentiates "ground-state" electron configurations. I know that we write the 3d^3 before 4s^2 because the d-orbital ends up being lower energy than the s orbital, so we have to wr...
by Kaylee Messick 3J
Sun Nov 01, 2020 6:36 pm
Forum: Student Social/Study Group
Topic: Sappling #25
Replies: 4
Views: 47

Re: Sappling #25

Hello! To find the energy of the photon, you would use E = hc/wavelength. To find the energy of an electron, you would use the equation for kinetic energy. To find kinetic energy, you will first need to find the velocity of the electron using the de Broglie wavelength equation. Hope this helps!
by Kaylee Messick 3J
Sun Nov 01, 2020 6:27 pm
Forum: Properties of Electrons
Topic: sapling no. 8
Replies: 4
Views: 93

Re: sapling no. 8

Hello! For this problem, you are trying to determine the energy levels (n). You already know the ending energy level because the problem states that blue light was emitted (visible light), which belongs to the Balmer series. Then to find the starting value of n, you could either use the Rydberg equa...
by Kaylee Messick 3J
Fri Oct 30, 2020 5:18 pm
Forum: DeBroglie Equation
Topic: Sapling HW #19
Replies: 2
Views: 39

Re: Sapling HW #19

I just realized my mistake was on one of the other questions on #19. Thank you for your reply though!
by Kaylee Messick 3J
Fri Oct 30, 2020 5:11 pm
Forum: DeBroglie Equation
Topic: Sapling HW #19
Replies: 2
Views: 39

Sapling HW #19

I keep getting this question incorrect. I used the DeBroglie equation, which is wavelength = h/(mass*velocity). I rearranged it to show velocity = h/(mass*wavelength). Then I put in the values (for proton) as velocity = (6.626*10^-34 J.s)/[(1.673*10^-27 kg) * (575.0 * 10^-9 m)] and I got 0.6888 m/s ...
by Kaylee Messick 3J
Sun Oct 25, 2020 9:39 pm
Forum: DeBroglie Equation
Topic: Sapling Hw 2 #19
Replies: 10
Views: 157

Re: Sapling Hw 2 #19

I also keep getting this question wrong. I used the DeBroglie equation, which is wavelength = h/(mass*velocity). I rearranged it to show velocity = h/(mass*wavelength). Then I put in the values (for proton) as velocity = (6.626*10^-34 J.s)/[(1.673*10^-27 kg) * (575.0 * 10^-9 m)] and I got 0.6888 m/s...
by Kaylee Messick 3J
Sun Oct 25, 2020 9:06 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Module #18
Replies: 5
Views: 56

Re: Module #18

I had a question about this problem too. Does it matter that the uncertainty in position is for the radius? I noticed several other problems use the diameter of the atom, so for this problem would we have to multiply the radius by 2 then find 1% or just calculate 1% of the radius?
by Kaylee Messick 3J
Sun Oct 25, 2020 6:59 pm
Forum: Molarity, Solutions, Dilutions
Topic: Question about Fundamentals G.25?
Replies: 4
Views: 70

Re: Question about Fundamentals G.25?

Sorry I didn't see this post until now! If you look in section G.4 of the textbook, Figure G.9 in the dilution section explains how the number of molecules goes down while moles stays the same. Basically, they are only looking at a specific area, and when diluted, the specific area will have less mo...
by Kaylee Messick 3J
Sun Oct 25, 2020 6:16 pm
Forum: Einstein Equation
Topic: question on homework #7
Replies: 6
Views: 61

Re: question on homework #7

The unit angstrom is 10^-10 m, so from 4.16 x 10^-7,which is the answer that you calculated, move the decimal over until it reaches 10^-10. Hope this helps!
by Kaylee Messick 3J
Sat Oct 24, 2020 9:17 pm
Forum: Molarity, Solutions, Dilutions
Topic: Question about Fundamentals G.25?
Replies: 4
Views: 70

Re: Question about Fundamentals G.25?

The moles of solute do stay the same when diluting a solution. For example, the equation c=n/V is the equation for molarity. When we increase the volume of the solution (V), the molarity (c) should go down because we are dividing by the volume, therefore making c smaller. However, the number of mole...
by Kaylee Messick 3J
Sat Oct 24, 2020 9:07 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Textbook Question 1B.25
Replies: 4
Views: 95

Re: Textbook Question 1B.25

Hello! The h bar means h/2pi as mentioned above. This is why in the Heisenberg Uncertainty Equation, the textbook listed it as greater than or equal to (1/2)(h bar), and we multiple (1/2) and (h/2pi) to get h/4pi as we usually write in the equation.
by Kaylee Messick 3J
Sun Oct 18, 2020 8:00 pm
Forum: Properties of Light
Topic: Work Function
Replies: 3
Views: 57

Re: Work Function

As mentioned in the previous response, the work function is the threshold energy (the energy required to remove an electron); however, I don't think there is a way to solve for it because this is what the photoelectric experiment was intending to find and they did this experimentally. I think the pr...
by Kaylee Messick 3J
Sun Oct 18, 2020 7:36 pm
Forum: Photoelectric Effect
Topic: Textbook Question 1A.11
Replies: 3
Views: 65

Re: Textbook Question 1A.11

The Balmer series, Lyman series, and Paschen series depend on the final energy level. When electrons drop down to different energy levels, they emit different wavelengths of light. These wavelengths fall in different locations on the spectrum, which is what classifies the series. For question 1A.11,...
by Kaylee Messick 3J
Sun Oct 18, 2020 7:12 pm
Forum: Properties of Electrons
Topic: Light Emission
Replies: 4
Views: 45

Re: Light Emission

Hello! I'm not exactly sure of the answer to your question, but I would guess that there is a correlation. The color of the light emitted depends on the wavelength of light as it leaves when the electrons drop down to a different energy level. The electrons in different atoms have very specific wave...
by Kaylee Messick 3J
Sun Oct 18, 2020 6:26 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Textbook Question 1A.13
Replies: 2
Views: 30

Re: Textbook Question 1A.13

Hello! In one of the workshops, the derived equation for the Rydberg formula was: frequency = -R [(1/n(i)^2) - (1/n(f)^2)]. The first n is the initial energy level (so it would be 2 in this problem) and the second n is the final energy level (so it would be 1 in this problem). After substituting the...
by Kaylee Messick 3J
Fri Oct 16, 2020 5:10 pm
Forum: SI Units, Unit Conversions
Topic: Converting from kJ/mol to J
Replies: 2
Views: 30

Converting from kJ/mol to J

In one of the Photoelectric Post-Module Assessment questions (#29), we had to convert the work function from kJ/mol to J. When converting from kJ/mol to J, we divide by Avogadro's constant to cancel out the moles, then we multiply by 1000 to convert kJ to J. I was wondering why we divide by Avogadro...
by Kaylee Messick 3J
Sun Oct 11, 2020 4:29 pm
Forum: Molarity, Solutions, Dilutions
Topic: Sapling HW Number 7
Replies: 12
Views: 138

Re: Sapling HW Number 7

Hello! The question gives the total mass of the mixture, so you would use the percentage given to find the mass of CaCl2. After subtracting the mass of CaCl2 from the total mass, the remaining mass would be the water since the two components of the mixture are CaCl2 and water. Hope this helps!
by Kaylee Messick 3J
Sun Oct 11, 2020 4:20 pm
Forum: Empirical & Molecular Formulas
Topic: Percent Yield in Sapling HW #10
Replies: 1
Views: 47

Re: Percent Yield in Sapling HW #10

Hello! When I was doing this problem, the Sapling feedback that I received was mass = density x volume. For example, under 2-butanone, it says d=0.81g/mL. Make sure that you convert the reactant to grams with this conversion before starting the problem, as this may be the issue. Hope this helps!
by Kaylee Messick 3J
Fri Oct 09, 2020 2:33 pm
Forum: Molarity, Solutions, Dilutions
Topic: Fundamentals G17 Part B
Replies: 1
Views: 38

Fundamentals G17 Part B

I'm having trouble starting G17 part b. The problem states, "(b) Determine the mass of CuSO4 ⋅ 5H20 that must be used to prepare 250 mL of 0.20M CuSO4(aq)." For part A, I used the formula c = n/V then converted to grams of solute. I was thinking for part b you would do something similar an...

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