Search found 111 matches

by Gwendolyn Hill 2F
Thu Mar 11, 2021 4:11 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Cell Diagrams: Platinum
Replies: 3
Views: 24

Cell Diagrams: Platinum

Hi everybody, I was wondering if someone could explain to me how to determine if a metal is conductive or not? In terms of the periodic table, which ones are conductive? Because I understand that if it is conductive and in its solid state in the rxn you don't need to include platinum there, but I ha...
by Gwendolyn Hill 2F
Sun Mar 07, 2021 3:01 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Textbook 6M.7
Replies: 1
Views: 23

Re: Textbook 6M.7

Hi! I had had this question as well haha.. basically, you use the reduction with the amount of electrons that most closely matches the other ones. For example, for a.), Zn has only one reduction possibility with Zn 2+ + 2e-, so you use the 2+ versions of the other ones. For d.), you use the closest ...
by Gwendolyn Hill 2F
Sun Mar 07, 2021 2:56 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: 7E.5 Question on Catalysts
Replies: 3
Views: 24

Re: 7E.5 Question on Catalysts

Hi! Following a general rule for determining which species are catalysts and intermediates helps me a lot. Both intermediates and catalysts do not appear in the overall reaction. A species is an intermediate if it is created in a reaction and then used up in a subsequent reaction. Because catalysts...
by Gwendolyn Hill 2F
Fri Mar 05, 2021 8:54 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: 7E.5 Question on Catalysts
Replies: 3
Views: 24

7E.5 Question on Catalysts

Hi,
Does anybody know how we identify which molecule in a reaction is the catalyst for reactions in general? In this problem OH- is the catalyst, and I was wondering how we were supposed to determine that since the OH- is canceled out in the rxn just like the intermediates.
by Gwendolyn Hill 2F
Wed Mar 03, 2021 12:34 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: Textbook 6N.5
Replies: 4
Views: 66

Textbook 6N.5

Hi, so I understand how to solve for the pH in part a. However, in the very last step in the solutions manual, we go straight from having 0.06V = -0.0257 V x (2.303log[H+]) to pH = 0.06V/(2.303 x 0.025693V). Where did the negative sign go? I keep getting -1.0 instead of 1.0 like they say because of ...
by Gwendolyn Hill 2F
Tue Mar 02, 2021 3:47 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Textbook 6M.7
Replies: 4
Views: 47

Re: Textbook 6M.7

Looking up the values in Appendix 2B was what I did. Essentially, you'd look for the values corresponding to the half reactions where the metal in its ionic state reduces back to its neutral state. But there are multiple where there is a plus3 state and then a plus2 state so which one do I use beca...
by Gwendolyn Hill 2F
Sun Feb 28, 2021 3:21 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Textbook 6M.7
Replies: 4
Views: 47

Re: Textbook 6M.7

Looking up the values in Appendix 2B was what I did. Essentially, you'd look for the values corresponding to the half reactions where the metal in its ionic state reduces back to its neutral state. But there are multiple where there is a plus3 state and then a plus2 state so which one do I use beca...
by Gwendolyn Hill 2F
Sun Feb 28, 2021 3:21 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Cell Diagram Question (again lol)
Replies: 2
Views: 18

Re: Cell Diagram Question (again lol)

I think a good rule of thumb would be to look at the phases of the reactants and products. If one of the phases is solid, then it's a conducting metal. If neither are solid, then they need another metal to act as its electrode. For example, in the question you mentioned, both Cr 2 O 7 2- and Cr 3+ ...
by Gwendolyn Hill 2F
Sun Feb 28, 2021 3:06 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Cell Diagram Question (again lol)
Replies: 2
Views: 18

Cell Diagram Question (again lol)

Hi! How do we know when to use platinum as a conducting electrode? I know that you use it when you are not using a conductive metal, but how would you be able to tell whether or not a metal is conductive? For example, Chromium from 6M.5 c.) ? Thanks!
by Gwendolyn Hill 2F
Sat Feb 27, 2021 8:22 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Cell Diagrams Question
Replies: 3
Views: 35

Cell Diagrams Question

Hi,
Would someone be able to explain to me how you decide to order the cell diagram? I know that in the textbook it says you put it in the order of "contact", but what does that mean? How do you figure any of this out? Thanks!
by Gwendolyn Hill 2F
Wed Feb 24, 2021 11:14 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Cell Diagrams
Replies: 1
Views: 48

Re: Cell Diagrams

Hi! We have to include Cl-, but we do not include K, even though we know K to be there. This is because Cl- does participate in the reactions like you said! We still need it for the reaction. Also, I believe the order in cell diagrams is to place the phases in order in which they are in contact. So ...
by Gwendolyn Hill 2F
Wed Feb 24, 2021 11:10 pm
Forum: Balancing Redox Reactions
Topic: Sapling 7/8 #18
Replies: 2
Views: 13

Re: Sapling 7/8 #18

Ditto what is said above! Also, make sure that you don't use parentheses around the product: Sapling is fickle that way....You also should not include any electrons, because in a balanced equation they cancel out!
by Gwendolyn Hill 2F
Sun Feb 21, 2021 6:31 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: calculating H through bond enthalpies
Replies: 4
Views: 19

Re: calculating H through bond enthalpies

Hi!

The negative sign described is due to the fact that the product bonds are being formed, and the reactant bonds are being broken. So when calculation total reaction enthalpy, you subtract the amount of energy for the products!
by Gwendolyn Hill 2F
Sun Feb 21, 2021 6:30 pm
Forum: Balancing Redox Reactions
Topic: Easy way to remember reduction/oxidazing agents?
Replies: 10
Views: 48

Re: Easy way to remember reduction/oxidazing agents?

Not sure I entirely understand the question about remembering, but it's not too difficult to figure out, I suppose? If you are looking for tips on figuring out which is which, I would remember OIL RIG (I know that Lavelle was saying LEO, etc). OIL: Oxidation Is Loss (of e-) and RIG: Reduction Is Gai...
by Gwendolyn Hill 2F
Sun Feb 21, 2021 6:28 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Hfus
Replies: 9
Views: 56

Re: Hfus

We use Hfus whenever we are trying to calculate a phase change from solid -> Liquid, or liquid-> solid (-Hfus). When trying to heat an ice cube and then eventually evaporate it, we would use Hfus first because of the phase change.
by Gwendolyn Hill 2F
Sun Feb 21, 2021 6:26 pm
Forum: Balancing Redox Reactions
Topic: Oxidation States for TM
Replies: 6
Views: 37

Re: Oxidation States for TM

Yeah, unless it's given. It's not too difficult to figure out if you can figure out the oxidation state of the atom(s) it is paired with in the equation!
by Gwendolyn Hill 2F
Sun Feb 21, 2021 6:25 pm
Forum: Ideal Gases
Topic: Calculating temperature change
Replies: 4
Views: 29

Re: Calculating temperature change

Hi! So because we have constant pressure, and we are given work, mols, and given delta U, we can calculate the temperature change by using the equation for q where q = nC∆T. We know it is an ideal gas, so I believe that we can say that Cp = 5/2 assuming it is monoatomic? Someone else can feel free t...
by Gwendolyn Hill 2F
Mon Feb 15, 2021 12:15 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Textbook 4A.13 Clarification
Replies: 2
Views: 41

Textbook 4A.13 Clarification

Hi! So for this question, I understand how to solve it; I just am getting a different answer than is in the solution manual. "A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solu- tion in the calorimeter (q 5 23.50 kJ), resu...
by Gwendolyn Hill 2F
Mon Feb 15, 2021 12:09 pm
Forum: Student Social/Study Group
Topic: Sapling Week 5-6 Homework Question #8
Replies: 2
Views: 25

Re: Sapling Week 5-6 Homework Question #8

Whoops! Sorry I mistyped; 298.15 was a final temp value haha... 273.15 is correct
by Gwendolyn Hill 2F
Mon Feb 15, 2021 12:08 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Sapling Week 5/6 #8
Replies: 6
Views: 53

Re: Sapling Week 5/6 #8

Hi! So for this, you actually just assume that the mols are 1. So the way I like to think about it is that this reaction has 4 steps. Step 1 is heating the reactants to 100 degrees Celsius,. So first, we have to convert our celsius to Kelvin by using the conversion 298.15K + degrees celsius given. ...
by Gwendolyn Hill 2F
Mon Feb 15, 2021 12:08 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Sapling number 8
Replies: 7
Views: 111

Re: Sapling number 8

Whoops! Sorry I mistyped; 298.15 was a final temp value haha... 273.15 is correct
by Gwendolyn Hill 2F
Sat Feb 13, 2021 8:35 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Sapling #8 week 5/6
Replies: 4
Views: 57

Re: Sapling #8 week 5/6

Hi! I think it is easiest to think about this problem in three steps (like the hint): 1. heating the water from 50.0 °C to its normal boiling point, 100.0 °C , I used the equation ∆S=nC ln(T2/T1) where C is the heat capacity of liquid water, T2=373 K and T1= 323 K (remember to convert celcius to Ka...
by Gwendolyn Hill 2F
Sat Feb 13, 2021 8:32 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Sapling number 8
Replies: 7
Views: 111

Re: Sapling number 8

Hi! So the way I like to think about it is that this reaction has 4 steps. Step 1 is heating the reactants to 100 degrees Celsius,. So first, we have to convert our celsius to Kelvin by using the conversion 298.15K + degrees celsius given. Next, we assume 1.00 mol, and use the equation ∆S = n*Cp*ln(...
by Gwendolyn Hill 2F
Sat Feb 13, 2021 8:31 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Sapling Week 5/6 #8
Replies: 6
Views: 53

Re: Sapling Week 5/6 #8

Hi! So for this, you actually just assume that the mols are 1. So the way I like to think about it is that this reaction has 4 steps. Step 1 is heating the reactants to 100 degrees Celsius,. So first, we have to convert our celsius to Kelvin by using the conversion 298.15K + degrees celsius given. N...
by Gwendolyn Hill 2F
Sat Feb 13, 2021 8:29 pm
Forum: Student Social/Study Group
Topic: Sapling Week 5-6 Homework Question #8
Replies: 2
Views: 25

Re: Sapling Week 5-6 Homework Question #8

Hi! So the way I like to think about it is that this reaction has 4 steps. Step 1 is heating the reactants to 100 degrees Celsius,. So first, we have to convert our celsius to Kelvin by using the conversion 298.15K + degrees celsius given. Next, we assume 1.00 mol, and use the equation ∆S = n*Cp*ln(...
by Gwendolyn Hill 2F
Sat Feb 13, 2021 8:26 pm
Forum: Student Social/Study Group
Topic: Sapling Week 5-6 Homework Question #5
Replies: 2
Views: 14

Re: Sapling Week 5-6 Homework Question #5

Hi! So first step, we have to convert T to units of Kelvin, so you take the values you are given and add them to 273.15. We have Tfinal and Tinitial, and are trying to find ∆S with constant volume Cv given, so that means we use the equation ∆S = n*Cv*ln(T2/T1). We are also given the pressure in kPa,...
by Gwendolyn Hill 2F
Sat Feb 06, 2021 4:38 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Week 3/4 Sapling #18
Replies: 4
Views: 46

Re: Week 3/4 Sapling #18

Hi! I had been having a hard time with this too. So basically you need to recognize that we are told Cp = 4R The relationship between Cp and Cv is: Cv = Cp-R. So 4R-R = 3R. ∆U = q + w. Then, understand that because we are at constant volume, ∆U = q + 0, since no change in volume means no work. q is...
by Gwendolyn Hill 2F
Sat Feb 06, 2021 3:34 pm
Forum: Calculating Work of Expansion
Topic: Sapling 15
Replies: 6
Views: 59

Re: Sapling 15

Hi! So first step, you have to convert the grams given of NaN3 to mols of N2. You use stoichiometry for this, and the ratios given in the reaction. You have to use the periodic table to find the molar mass of NaN3. Second step, N2 acts as an ideal gas, so you can use the ideal gas equation, PV = nRT...
by Gwendolyn Hill 2F
Sat Feb 06, 2021 3:31 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Sapling Week 3 and 4, Q5
Replies: 4
Views: 28

Re: Sapling Week 3 and 4, Q5

Hi! So for this reaction, we are given three equations: Consider these reactions, where M represents a generic metal. 2M(s)+6HCl(aq)⟶2MCl3(aq)+3H2(g) ΔH1=−787.0 kJ HCl(g)⟶HCl(aq) ΔH2=−74.8 kJ H2(g)+Cl2(g)⟶2HCl(g) ΔH3=−1845.0 kJ MCl3(s)⟶MCl3(aq) ΔH4=−376.0 kJ Use the given information to determine th...
by Gwendolyn Hill 2F
Sat Feb 06, 2021 3:26 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: sapling #18
Replies: 2
Views: 26

Re: sapling #18

Oops! Sorry that was for the second half of the problem with ∆U.
To find q, you need to use the same equation:

q = n * C * ∆T

and just plug in your values
by Gwendolyn Hill 2F
Sat Feb 06, 2021 3:26 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: sapling #18
Replies: 2
Views: 26

Re: sapling #18

Hi! I had been having a hard time with this too. So basically you need to recognize that we are told Cp = 4R The relationship between Cp and Cv is: Cv = Cp-R. So 4R-R = 3R. ∆U = q + w. Then, understand that because we are at constant volume, ∆U = q + 0, since no change in volume means no work. q is ...
by Gwendolyn Hill 2F
Sat Feb 06, 2021 3:24 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Week 3/4 Sapling #18
Replies: 4
Views: 46

Re: Week 3/4 Sapling #18

Hi! I had been having a hard time with this too. So basically you need to recognize that we are told Cp = 4R The relationship between Cp and Cv is: Cv = Cp-R. So 4R-R = 3R. ∆U = q + w. Then, understand that because we are at constant volume, ∆U = q + 0, since no change in volume means no work. q is ...
by Gwendolyn Hill 2F
Sat Feb 06, 2021 3:20 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Sapling question
Replies: 3
Views: 28

Re: Sapling question

Hi! The hint given is: The change in entropy of the system is given by ΔS∘sys=∑S∘(products)−∑S∘(reactants) where S∘ is the standard entropy. Take into account the coefficients from the balanced chemical equation. It says refer to the standard entropy values as needed: so you can click on that and fi...
by Gwendolyn Hill 2F
Sun Jan 31, 2021 9:47 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Cp & Cv
Replies: 7
Views: 62

Re: Cp & Cv

HI! So Cv is the molar heat capacity of a gas at constant volume and Cp is molar heat capacity of a gas at constant pressure. Increased temperature of a gas causes the molecules to be more excited---> tries to expand in volume. If you stop the piston from moving, you can use the Cv, since then there...
by Gwendolyn Hill 2F
Sun Jan 31, 2021 9:31 pm
Forum: Calculating Work of Expansion
Topic: assuming room temperature
Replies: 4
Views: 36

Re: assuming room temperature

Yes, I believe that is the case! I would try calculating it with room temperature, checking it against the textbook answers, and then if you got it wrong then try again. You shouldn't be expected to like.... divine the temperature of a random room with absolutely no context lol.
by Gwendolyn Hill 2F
Sun Jan 31, 2021 9:25 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Sapling #15
Replies: 3
Views: 29

Re: Sapling #15

I did the same as the people above; just make sure that when you use R, you are using the correct one with the correct units: This one, I believe you should use R = 0.0821 L·atm/(mol·K)
by Gwendolyn Hill 2F
Sun Jan 31, 2021 9:19 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Temperature Change
Replies: 14
Views: 92

Re: Temperature Change

Hi!

An Endothermic reaction is defined as absorbing a net amount of heat, and an increase in temperature causes the reaction to favor product formation. Because the reaction starts to favor product formation, K will become a larger value (products over reactants is how you calculate K).
by Gwendolyn Hill 2F
Sun Jan 31, 2021 9:07 pm
Forum: Calculating Work of Expansion
Topic: Which R to use?
Replies: 4
Views: 289

Re: Which R to use?

Hi! You always use the R that matches the unit: in this case, R = 0.0821 L·atm/(mol·K)
by Gwendolyn Hill 2F
Sun Jan 24, 2021 4:24 pm
Forum: Phase Changes & Related Calculations
Topic: Ways to calculate enthalpy
Replies: 4
Views: 51

Re: Ways to calculate enthalpy

Ditto with everything above! I would also say that for the third method, with the standard enthalpy of the whole reaction, we should be given the values necessarily (the enthalpies of each of the parts)!
by Gwendolyn Hill 2F
Sun Jan 24, 2021 4:16 pm
Forum: Ideal Gases
Topic: How can we identify when to use the ideal gas law?
Replies: 11
Views: 67

Re: How can we identify when to use the ideal gas law?

As said above, if a problem gives you multiple elements of the ideal gas law and asks you to solve for one of them, like for pressure for example, then you would use the law. That is also especially true if you're not given a Kp; it means you can't do the ICE table right off the bat.
by Gwendolyn Hill 2F
Sun Jan 24, 2021 4:12 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Standard Enthalpy Chart
Replies: 2
Views: 36

Re: Standard Enthalpy Chart

We should be given all information necessary in order to solve these. We're not expected to memorize them all and store em in our brain :)
by Gwendolyn Hill 2F
Sat Jan 23, 2021 5:06 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: Textbook 6D.5
Replies: 2
Views: 45

Re: Textbook 6D.5

Peter DePaul 1E wrote:Yeah you would have to find the value from the textbook and whatever value they provide is the one to use.

Thanks!
by Gwendolyn Hill 2F
Sat Jan 23, 2021 4:38 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: Textbook 6D.5
Replies: 2
Views: 45

Textbook 6D.5

Hi! So the problem is as follows: 6D.5 Calculate the pH, pOH, and percentage protonation of sol- ute in each of the following aqueous solutions: (a) 0.057 m NH3(aq); (b) 0.162 m NH2OH(aq); (c) 0.35 m (CH3)3N(aq); (d) 0.0073 m codeine, given that the pKa of its conjugate acid is 8.21. On part a, in t...
by Gwendolyn Hill 2F
Sat Jan 23, 2021 4:34 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Textbook Problem 6D.3
Replies: 1
Views: 14

Re: Textbook Problem 6D.3

Hi!
This is actually a problem with sigfigs; you can see that in the solutions manual they used (0.06)^2/(0.10-0.06), which actually does give 0.09. I actually did the exact same thing as you and was like wait a minute.... haha. So it's just sig figs!
by Gwendolyn Hill 2F
Sun Jan 17, 2021 2:01 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 5.35
Replies: 1
Views: 9

Re: 5.35

Hi!

So it seems like the value that C levels off at is 10, and B levels off at 5, so C is twice the amount of B in terms of partial pressure. If A had a coefficient of 1, I believe we would see it level off closer to C and B, and A looks ~ 2x higher on the chart.
by Gwendolyn Hill 2F
Sun Jan 17, 2021 1:58 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: ICE Chart with Gas Pressures
Replies: 8
Views: 65

Re: ICE Chart with Gas Pressures

Hi!
So for the sake of this class, Lavelle said that we are using 1 bar = 1 atm, so you can just kind of "say" they are equal. And yes! That is how you would do the ICE chart. Use whatever version of the compounds you are given unless stated otherwise!
by Gwendolyn Hill 2F
Sun Jan 17, 2021 1:38 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Sapling Question
Replies: 4
Views: 36

Re: Sapling Question

Hi! So you are correct that the change for both reactants would be +1 while the change for the product would be -2x. However, the initial concentrations of the reactants should be .1, because you were at equilibrium and it was disturbed, so you need to find the new equilibrium values. So the initial...
by Gwendolyn Hill 2F
Sun Jan 17, 2021 1:31 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Textbook Problem 5I.19
Replies: 2
Views: 17

Re: Textbook Problem 5I.19

Hi! I believe how you would account for the 60% part is that 60% of H2 is now "gone" (reacted with I2 to form HI on the other side), so at equilibrium you are left with 40% of the original amount. In your ICE box, this would mean you started with 0.400 mol / 3.00 L of H2 (which is the same...
by Gwendolyn Hill 2F
Tue Jan 12, 2021 2:41 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Sapling Homework #5
Replies: 2
Views: 49

Re: Sapling Homework #5

Thank you so much!
by Gwendolyn Hill 2F
Tue Jan 12, 2021 2:16 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Sapling Homework #5
Replies: 2
Views: 49

Sapling Homework #5

Hello, I am unsure how to solve this problem: Consider the reaction of NH3 and I2 to give N2 and HI. 2NH3(g)+3I2(g)↽−−⇀N2(g)+6HI(g). K? Using two or more of the given equations, determine the equilibrium constant, K , for the reaction of NH3 with I2. H2(g)+I2(g)↽−−⇀2HI(g) Ka=160 I2(g)↽−−⇀2I(g) Kb=2....
by Gwendolyn Hill 2F
Sat Jan 09, 2021 11:38 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Does K refer to Kc or Kp or both?
Replies: 4
Views: 55

Re: Does K refer to Kc or Kp or both?

K is the general name for the constant! When talking about measuring concentration, we use Kc, and for pressure (atm or bars) we use Kp. Kc is measured in molarity (concentration) of the reactants and products, and Kp is measured in the partial pressure P. I think what you're saying should be correc...
by Gwendolyn Hill 2F
Sat Jan 09, 2021 11:35 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Q and K
Replies: 13
Views: 61

Re: Q and K

Everything above is true: I would also just add on that if we are not sure that the reaction is at equilibrium but we are given K, I believe we can also have Q = K and prove the reaction is at equilibrium, since the process for finding Q is the same as the process for finding K. :)
by Gwendolyn Hill 2F
Sat Jan 09, 2021 11:33 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Products and Equilibrium
Replies: 5
Views: 47

Re: Products and Equilibrium

More products at equilibrium refers to the concentration of products, which as stated above needs to be understood as K > 1.0x10^3 in order for the reaction to favor the products. Otherwise, there might just be a higher concentration of product, but if it is in that intermediate value between 1.0x10...
by Gwendolyn Hill 2F
Sat Jan 09, 2021 11:31 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Calculating Q
Replies: 8
Views: 46

Re: Calculating Q

Q is used to check reactions to see if they are equilibrium, and Q is the "reaction quotient" of a time that is "unknown". Q is calculated in the exact manner that K is calculated. Q<K means R->P, Q>K means R<-P. Q = K means you are at equilibrium.
by Gwendolyn Hill 2F
Sat Jan 09, 2021 11:30 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Reactants and equilibrium
Replies: 3
Views: 39

Re: Reactants and equilibrium

As said above, more reactants at equilibrium just means a higher concentration of reactants, and the region mentioned of 1.0x10^-3 < K < 1.0x10^3 means neither R or P are truly favored to the extent where it is important. When there are significantly more of the reactants, the equilibrium will sit t...
by Gwendolyn Hill 2F
Wed Dec 16, 2020 11:44 am
Forum: Amphoteric Compounds
Topic: More likely acid or base
Replies: 4
Views: 78

Re: More likely acid or base

Amphoteric compounds will react to their environments accordingly, like you guessed. For HCO3-, for example, it can turn into CO3(2-) or H2CO3, depending on the balance of the pH of the solution it is in. If HCO3- is combined with an acid, it would act like a base.
by Gwendolyn Hill 2F
Wed Dec 16, 2020 11:41 am
Forum: Polyprotic Acids & Bases
Topic: Solving for polyprotic acids and bases
Replies: 4
Views: 79

Re: Solving for polyprotic acids and bases

I think so; this method seems to work to organize the equations and show the ionization stepwise. For example, polyprotic acids like H3PO4, H2PO4-, and HPO4(2-) will always show that each ionization step is more difficult, since it's more difficult to remove H+ from a molecule as its negative charge...
by Gwendolyn Hill 2F
Wed Dec 16, 2020 11:37 am
Forum: Acidity & Basicity Constants and The Conjugate Seesaw
Topic: What is the Conjugate Seesaw
Replies: 11
Views: 101

Re: What is the Conjugate Seesaw

I believe that the conjugate see-saw concept is about the stronger the acid, the weaker the conjugate base, and the stronger the base, the weaker the conjugate acid. Like a seesaw! The stronger acid will "weigh down" their side of the seesaw, and therefore the base will be "higher up&...
by Gwendolyn Hill 2F
Wed Dec 16, 2020 11:31 am
Forum: *Indicators
Topic: Which indicator?
Replies: 2
Views: 62

Re: Which indicator?

If you're talking about the paper indicators, the strips you use to test pH, then?? Or do you mean stuff like soil, flower petals, leaves?? If you're talking about indicators as weak acids, I would imagine that using an indicator that falls in the middle of the range would work... or for a more effe...
by Gwendolyn Hill 2F
Sat Dec 12, 2020 8:54 pm
Forum: Conjugate Acids & Bases
Topic: Textbook 6.21
Replies: 1
Views: 65

Textbook 6.21

Hi, I had a question about textbook problem 6.21. The problem is as follows: "The two strands of the nucleic acid DNA are held together by hydrogen bonding between four organic bases. The structure of one of these bases, thymine, is shown below. (a) How many pro- tons can this base accept? (b) ...
by Gwendolyn Hill 2F
Sun Dec 06, 2020 9:02 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Textbook 2E13C
Replies: 3
Views: 65

Re: Textbook 2E13C

Pretty sure that since I is a larger molecule, the single bond representation is more accurate. I'm sure that having resonance structures combining would give us a bond length somewhere in between, but it might be because single bonds are longer and I is a larger molecule, and double bonds are short...
by Gwendolyn Hill 2F
Sun Dec 06, 2020 9:00 pm
Forum: Hybridization
Topic: Hybridization Format
Replies: 2
Views: 40

Re: Hybridization Format

Yes! It IS because you are using level 2 s and p orbitals. If you had a different molecule with hybridized orbitals and it had s and p hybridization but it's a larger molecule, then it would go up in level, like 3sp^2 or etc
by Gwendolyn Hill 2F
Sun Dec 06, 2020 8:56 pm
Forum: Shape, Structure, Coordination Number, Ligands
Topic: Coordination Number
Replies: 4
Views: 56

Re: Coordination Number

We might have to know them.... they're on the list of coordination compounds from his website;
https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14A/NamingCoordinationCompounds.pdf
by Gwendolyn Hill 2F
Sun Dec 06, 2020 8:53 pm
Forum: Lewis Acids & Bases
Topic: CH4 and NH3
Replies: 3
Views: 19

Re: CH4 and NH3

CH4, methane, is an extremely stable molecule; in order for something to be an acid or a base (or act as one), the molecule needs to be an easy electron donor or recipient. The hydrogen atoms are too tightly held to dissociate in water. For NH3, it is a substance that accepts an H+ from water so nor...
by Gwendolyn Hill 2F
Sun Dec 06, 2020 8:49 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Sapling Week 9 HW #5
Replies: 5
Views: 59

Re: Sapling Week 9 HW #5

Yea I thought it was kind of unfair that we were expected to know that shorthand. EN stands for Ethylenediamine, which is H2NCH2CH2NH2. It's bidentate, so it contributes two to the coordination number. So then 2 x 2 = 4 for en, and then 2 more CO, so it should be six?
by Gwendolyn Hill 2F
Sun Dec 06, 2020 8:46 pm
Forum: Hybridization
Topic: Pi bonds Hybridization
Replies: 2
Views: 24

Re: Pi bonds Hybridization

Not sure what you mean; It's just when they overlap. In a previous post on this website, someone said "The purpose of hybridization is to see the types of bonds that the atoms share with each other, whether it be sigma or pi bonds. The different types of bonds allow different properties, like h...
by Gwendolyn Hill 2F
Sun Dec 06, 2020 8:41 pm
Forum: Sigma & Pi Bonds
Topic: Pi bonds
Replies: 7
Views: 63

Re: Pi bonds

Generally you just know that 1 of them is a pi bond in a double bond, and Lavelle draws the triple bonds with two pi bonds on the outside. It's just a way of thinking about them, since in reality the pi bonds are in a general "space" (lewis structures aren't really super accurate), and the...
by Gwendolyn Hill 2F
Sun Nov 29, 2020 11:36 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Remembering Bond Degrees
Replies: 7
Views: 116

Re: Remembering Bond Degrees

Visualization helps me a lot! If you can remember how many electron regions there are and the different types, then you're basically just doing variations on the same model. Fortunately, we are not expected to know exACTLY what the bond angles are for the weird ones which have lots of lone electron ...
by Gwendolyn Hill 2F
Sun Nov 29, 2020 11:34 pm
Forum: Hybridization
Topic: Delocalized Pi Bond
Replies: 10
Views: 98

Re: Delocalized Pi Bond

Delocalized pi bonds are pi bonds where the e- are free to move over more than two nuclei, like the example of Benzene that Prof. Lavelle gave. Delocalized pi bonds can also be thought of as where there are pie bonds with resonance structures, since the lewis structures we draw are not entirely accu...
by Gwendolyn Hill 2F
Sun Nov 29, 2020 11:31 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Tetrahedral molecular shape
Replies: 10
Views: 138

Re: Tetrahedral molecular shape

Ditto what is said above; but also, you just have to remember that in order for the four atoms to be the furthest apart, 90 degrees isn't enough. If you're thinking in an XYZ plane, 90 degrees is what you'd get for octahedral geometry
by Gwendolyn Hill 2F
Sun Nov 22, 2020 8:42 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: sapling #4
Replies: 3
Views: 18

Re: sapling #4

He also went over the general formula, with the whole A = central atom X=bonded atoms E=lone pairs thing, and if multiple molecules have the same AXE formula, they have the same shape :)
by Gwendolyn Hill 2F
Sun Nov 22, 2020 8:39 pm
Forum: Hybridization
Topic: Hybridization and Covalent Bonds
Replies: 2
Views: 26

Re: Hybridization and Covalent Bonds

I'm like 90% sure that's correct, since ionic bonds involve a transfer; ionic bonds are ONLY electrostatic attraction, and true ionic bonds aren't really directional, so there's not much geometry to them either
by Gwendolyn Hill 2F
Sun Nov 22, 2020 8:37 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: linear shape?
Replies: 5
Views: 27

Re: linear shape?

What Julianna said! And also, if you're talking specifically about truly linear shapes, this will only happen when there are not any lone pairs to distort the molecule. For example, H2O has two lone pairs on the central atom (O), which distorts the shape and makes it "Bent" or "Angula...
by Gwendolyn Hill 2F
Sun Nov 15, 2020 6:41 pm
Forum: Ionic & Covalent Bonds
Topic: bond lengths
Replies: 9
Views: 108

Re: bond lengths

Double bonds are shorter than single bonds, because the strength of a bond is measured by its dissociation energy; the energy required to break a bond. Double bonds are stronger because they are more localized to the nucleus where the effective nuclear charge is.
by Gwendolyn Hill 2F
Sun Nov 15, 2020 6:40 pm
Forum: Lewis Structures
Topic: Sapling Weeks 5/6 HW #18
Replies: 7
Views: 65

Re: Sapling Weeks 5/6 HW #18

Basically, you can just think about it in terms of electrons too if that makes it easier. When molecules are larger, their radius increases and the valence electrons are less tightly held. This means that their electron density "cloud" is larger, and the electrons are more easily influence...
by Gwendolyn Hill 2F
Sun Nov 15, 2020 6:34 pm
Forum: Bond Lengths & Energies
Topic: Hydrogen Bonds
Replies: 16
Views: 195

Re: Hydrogen Bonds

No; Hydrogen bonding occurs because hydrogen is "electropositive", or only slightly electronegative. N, O, and F are highly electronegative, and thus attract hydrogen. Hydrogen doesn't have enough strength to attract other hydrogens' electrons if they are already separately bonded
by Gwendolyn Hill 2F
Sun Nov 15, 2020 6:18 pm
Forum: Ionic & Covalent Bonds
Topic: Textbook problem
Replies: 1
Views: 16

Re: Textbook problem

Hi! So remember that ionic bonds can be identified as a bond between a cation and an anion, like Na+ and Cl-. To find out how much ionic character a bond shows, you look at the electron affinity of each atom, and the electronegativity. Usually in ionic bonds, each atom gains or loses an electron. In...
by Gwendolyn Hill 2F
Sun Nov 15, 2020 6:15 pm
Forum: Lewis Structures
Topic: Expanded Valence Shells
Replies: 2
Views: 17

Re: Expanded Valence Shells

Hi! So, basically, the exceptions to the octet guideline can basically be summarized as elements in row 3+ in the P block. It is because they have d-orbitals in the valence shell, and those can accomodate additional e-. Another exception are Radicals; compounds with unpaired electrons that make it r...
by Gwendolyn Hill 2F
Sun Nov 15, 2020 6:12 pm
Forum: Ionic & Covalent Bonds
Topic: Difference on how atomic size affects covalent and dipole interactions
Replies: 9
Views: 66

Re: Difference on how atomic size affects covalent and dipole interactions

Yes, that is basically correct; the thing to remember is that for london dispersion forces, it's more about how since the electrons in their respective places are further away from the nuclear charge, the electron density area expands since they're less tightly held. This causes the electrons to be ...
by Gwendolyn Hill 2F
Sun Nov 15, 2020 6:07 pm
Forum: Lewis Structures
Topic: Number of Bonds for Elements
Replies: 3
Views: 23

Re: Number of Bonds for Elements

Usually we follow the octet guideline, but elements in Row 3+ (P block) can have expanded octets. This can happen because atoms in period 3 or higher have d-orbitals in the valence shell that can accomodate additional e-. Ex: P, S, Cl can accomodate more than 8 valence e-. Remember that the octet gu...
by Gwendolyn Hill 2F
Sun Nov 15, 2020 6:05 pm
Forum: Bond Lengths & Energies
Topic: Interaction Potential Energy
Replies: 2
Views: 36

Re: Interaction Potential Energy

Interaction Potential Energy is the energy due to position, composition, or arrangement. Also, it is the energy associated with forces of attraction and repulsion between objects. Any object that is lifted from its resting position has stored energy therefore it is called potential energy because it...
by Gwendolyn Hill 2F
Sun Nov 15, 2020 6:00 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: PCL5 Trigonal planar
Replies: 3
Views: 20

Re: PCL5 Trigonal planar

Exactly what is said above; The most favorable position is when the atoms are the farthest away from each other because of repulsion, so that is when it is the most stable. Having an atom on top and one on bottom makes a 90 degree angle between all of the atoms
by Gwendolyn Hill 2F
Sun Nov 15, 2020 5:58 pm
Forum: Bond Lengths & Energies
Topic: Hydrogen Bonding
Replies: 13
Views: 77

Re: Hydrogen Bonding

Not that we'll need to look at I don't think. I looked it up, and the only reason that hydrogen bonding would happen without a N, O, or F molecule is: "C-H bonds only participate in hydrogen bonding when the carbon atom is bound to electronegative substituents as is the case in chloroform, CHCl...
by Gwendolyn Hill 2F
Sun Nov 15, 2020 5:56 pm
Forum: Ionic & Covalent Bonds
Topic: Sapling Q: Hydrogen Bonds
Replies: 7
Views: 73

Re: Sapling Q: Hydrogen Bonds

So the most electronegative elements, the N, O, F elements, are what we are looking for. We are also looking for hydrogens, of course, but basically you can just look for N, O, or F. Hydrogen is the least electronegative, so it is attracted to the highly electronegative molecules N, O, F. Also, ther...
by Gwendolyn Hill 2F
Sun Nov 15, 2020 5:51 pm
Forum: Dipole Moments
Topic: Do london dispersion forces occur for polar molecules
Replies: 3
Views: 23

Re: Do london dispersion forces occur for polar molecules

I think what it's basically saying is that when polar molecules interact, their main interaction is Dispersion forces are present between all molecules, whether they are polar or nonpolar. Larger and heavier atoms and molecules exhibit stronger dispersion forces than smaller and lighter ones. In a l...
by Gwendolyn Hill 2F
Sun Nov 08, 2020 11:55 am
Forum: Lewis Structures
Topic: Lone Pairs Question
Replies: 22
Views: 128

Re: Lone Pairs Question

Adding on to what Carolina said;
Lone pairs are important to show, because you have to represent all of the electrons present, bonded or un-bonded. These lone pairs are what can skew the structure of a molecule, a/o allow for bonding with other molecules that also have lone pairs :)
by Gwendolyn Hill 2F
Sun Nov 08, 2020 11:52 am
Forum: Quantum Numbers and The H-Atom
Topic: Principle Quantum #
Replies: 3
Views: 56

Re: Principle Quantum #

Hi! So quantum numbers themselves: A quantum number is a value that is used when describing the energy levels available to atoms and molecules. In order to describe an electron, four quantum numbers are needed; energy (n), angular momentum (ℓ), magnetic moment (mℓ), and spin (ms). n is basically the...
by Gwendolyn Hill 2F
Sun Nov 08, 2020 11:47 am
Forum: Ionic & Covalent Bonds
Topic: Question about bond Length
Replies: 6
Views: 72

Re: Question about bond Length

The length of the bond is determined by the number of bonded electrons (the bond order). The higher the bond order, the stronger the pull between the two atoms and the shorter the bond length. Generally, the length of the bond between two atoms is approximately the sum of the covalent radii of the t...
by Gwendolyn Hill 2F
Sun Nov 08, 2020 11:46 am
Forum: Resonance Structures
Topic: Question about Resonance Def
Replies: 5
Views: 50

Re: Question about Resonance Def

Resonance is a blending of structures that have the same arrangement of atoms, but different arrangements of e-. The combination of all of the resonance structures is what results in the actual composition of the molecule. Dr. Lavelle talked about how for nitrate, each N-O bond has partial double bo...
by Gwendolyn Hill 2F
Sun Nov 08, 2020 11:44 am
Forum: Administrative Questions and Class Announcements
Topic: CHEM 14B Class Structure
Replies: 2
Views: 32

Re: CHEM 14B Class Structure

Essentially, Chem 14A is asynchronous too since it's been all prerecorded. I assume 14B will be the same, especially because it is also going to be taught by Lavelle. I expect that Lavelle will teach 14B and then 14C in the spring, because professors normally follow their class through the series :)...
by Gwendolyn Hill 2F
Sun Nov 01, 2020 2:01 pm
Forum: Einstein Equation
Topic: E=hv vs E=hf
Replies: 15
Views: 139

Re: E=hv vs E=hf

I personally prefer to use f when I'm writing down problems, just because I don't want to get mixed up between the v for frequency and the V for velocity
by Gwendolyn Hill 2F
Sun Nov 01, 2020 1:59 pm
Forum: Properties of Electrons
Topic: Sapling HW #1
Replies: 6
Views: 57

Re: Sapling HW #1

wavelength and frequency do not change based on the number of photons there are, but 100 photons of energy E creates 100E, at that SAME wavelength and frequency :) Proportionality doesn't matter here, it's just kind of a check for understanding about how photons work.
by Gwendolyn Hill 2F
Sun Nov 01, 2020 1:38 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: Angular Momentum
Replies: 5
Views: 34

Re: Angular Momentum

Not sure if that is an O or a 0, but a value of 0 would mean that its linear momentum is zero. For an electron for example, that means the electron is stationary. For S orbitals, the angular momentum is 0, because S orbitals are spherically symmetric, and angular momentum is about irregular shapes o...
by Gwendolyn Hill 2F
Sun Nov 01, 2020 1:33 pm
Forum: Trends in The Periodic Table
Topic: Increasing effective nuclear charge across a period
Replies: 3
Views: 27

Re: Increasing effective nuclear charge across a period

So basically, the effective nuclear charge is the net positive charge experienced by valence electrons. The effective nuclear charge of an atom increases as the number of protons in an atom increases, and in terms of the electron part; Across a period, effective nuclear charge increases as electron ...
by Gwendolyn Hill 2F
Sun Nov 01, 2020 1:29 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: Sapling #24
Replies: 16
Views: 80

Re: Sapling #24

Like everyone else said, the wave has to connect to itself. If you start with a curve "above" the "x axis", you have to end with a curve coming up from "below" the "x axis". I think about it like that; how it connects to being a sin wave. :)
by Gwendolyn Hill 2F
Sat Oct 24, 2020 2:53 pm
Forum: SI Units, Unit Conversions
Topic: Midterm
Replies: 26
Views: 276

Re: Midterm

It seems like the longer questions will be 10 points, so they're pretty fat, but there will be shorter ones that are more conceptual or have only like 1 or 2 steps to the equations that will act as a buffer for those harder ones!
by Gwendolyn Hill 2F
Sat Oct 24, 2020 2:51 pm
Forum: Properties of Electrons
Topic: Rydberg
Replies: 10
Views: 129

Re: Rydberg

As long as you know which one is the final and which one is the initial, it shouldn't matter what notation you use :) it all essentially means the same thing, you just mostly have to understand how the electron is moving.
by Gwendolyn Hill 2F
Sat Oct 24, 2020 2:48 pm
Forum: Student Social/Study Group
Topic: Additional Textbook Practice Problems
Replies: 2
Views: 24

Re: Additional Textbook Practice Problems

I don't think you need to any textbook problems outside of the ones he assigned; he already assigned SO many, and he wouldn't throw something at us that he hadn't shown us before. Laos, sometimes he takes problems from the textbook that he had assigned to use on tests, so I think it makes more sense...
by Gwendolyn Hill 2F
Sat Oct 24, 2020 2:46 pm
Forum: Administrative Questions and Class Announcements
Topic: Questions on midterm
Replies: 7
Views: 43

Re: Questions on midterm

Since the midterm is only 50 minutes long, which is not much time, there shouldn't be too many super hard calculation problems. What I heard from one of the UA meetings was that he plans to throw in a bunch of smaller easier questions with either simpler calculations a/o conceptual questions to act ...
by Gwendolyn Hill 2F
Sat Oct 24, 2020 2:42 pm
Forum: Student Social/Study Group
Topic: Study tips?
Replies: 9
Views: 94

Re: Study tips?

I'm honestly sad that there won't be any more prerecorded lectures on his own website after these outlines, because for me they have been super helpful to watch before his lectures that he posts on CCLE. Those assessment surveys have been where I have done some practice, and I've been doing some tex...

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