Search found 109 matches

by JasonLiu_2J
Sun Mar 15, 2020 1:52 pm
Forum: Administrative Questions and Class Announcements
Topic: Athena
Replies: 34
Views: 1632

Re: Athena

Thanks for everything over the past two quarters Dr. Lavelle! Hope you have a nice break! *high five*
by JasonLiu_2J
Wed Mar 11, 2020 10:56 am
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: q vs delta H to calculate delta S
Replies: 1
Views: 39

Re: q vs delta H to calculate delta S

Remember that under conditions of constant pressure, delta H equals q. I think in this situation, they simply substituted in delta H for q in the change in entropy equation. Whether you call the heat generated delta H or q in this question does not matter, since it will still give you the same chang...
by JasonLiu_2J
Wed Mar 11, 2020 10:49 am
Forum: Reaction Mechanisms, Reaction Profiles
Topic: 7.23
Replies: 2
Views: 21

Re: 7.23

I think that was probably a mistake in the answer key because since Cl- does not cancel when combining the elementary steps, it should show up in the final balanced equation.
by JasonLiu_2J
Wed Mar 11, 2020 10:47 am
Forum: Administrative Questions and Class Announcements
Topic: Saying Thank You to Dr. Lavelle
Replies: 275
Views: 116739

Re: Saying Thank You to Dr. Lavelle

Dear Dr. Lavelle, Thank you so much for everything you have done for us over the past two quarters. When I first came to UCLA last fall, I did not know what to expect from your classes and was very nervous when I stepped into your class. But I can say without a doubt that having you as my Chem 14A a...
by JasonLiu_2J
Wed Mar 11, 2020 10:23 am
Forum: Administrative Questions and Class Announcements
Topic: ENDGAME Review Session
Replies: 71
Views: 2168

Re: ENDGAME Review Session

Thank you so much for everything Lyndon! You've been such a great UA and helped everyone so much throughout both Chem 14A and 14B. We really appreciate all the time you have put in to helping us students succeed. Good luck with your future endeavors after college as well!
by JasonLiu_2J
Tue Mar 10, 2020 3:24 pm
Forum: First Order Reactions
Topic: determining Kr
Replies: 5
Views: 28

Re: determining Kr

Also to add on, remember that for a first order reaction, the slope of the graph ln[A] vs time is -k, so you would need to multiply it by negative 1 to find the value of the rate constant.
by JasonLiu_2J
Fri Mar 06, 2020 12:18 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: 6N.11
Replies: 1
Views: 52

Re: 6N.11

The solubility product is essentially the equilibrium constant of the solubility expression for Hg2Cl2. When Hg2Cl2 dissolves in solution, you get the equation Hg2Cl2(s) --> Hg2^2+(aq) + 2Cl-(aq). Looking at appendix 2B, you find a reduction half reaction with the form Hg2Cl2(s) --> 2Hg(l) + 2Cl-(aq...
by JasonLiu_2J
Fri Mar 06, 2020 12:10 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 6L.7
Replies: 1
Views: 44

Re: 6L.7

I think for this question, you have to look in Appendix 2B at the different half reactions and see which ones can be incorporated together to give you the overall equation given. In (a) for example, you would look in the appendix and find the reduction half reaction AgBr(s) +e- ---> Ag(s) + Br-(aq) ...
by JasonLiu_2J
Fri Mar 06, 2020 12:03 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: Initial Rates
Replies: 3
Views: 52

Re: Initial Rates

If the question is asking to solve for k and n using through initial rates, then you would have to have concentrations and initial reaction rates given to you so that you can find k and n by comparing how changes in concentration affect the changes in initial reaction rates.
by JasonLiu_2J
Fri Mar 06, 2020 12:00 pm
Forum: Interesting Applications: Rechargeable Batteries (Cell Phones, Notebooks, Cars), Fuel Cells (Space Shuttle), Photovoltaic Cells (Solar Panels), Electrolysis, Rust
Topic: 6O.1
Replies: 2
Views: 63

Re: 6O.1

I believe that since you are electrolyzing the NiSO4(aq) solution, you are trying to force the reduction of Ni2+, as the Ni2+ reduction potential is negative. Since you are only given that NiSO4 is electrolyzed, you can assume that the other reaction will involve the oxidation of water to reduce Ni2...
by JasonLiu_2J
Tue Mar 03, 2020 3:31 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: Reducing Power
Replies: 3
Views: 36

Re: Reducing Power

Reducing power is the ability of a molecule/substance to reduce another. In a redox couple, this would be the reducing agent. When this reducing agent reduces another element, it itself becomes oxidized. Typically, those with a greater reducing power are more likely to donate their electrons to redu...
by JasonLiu_2J
Fri Feb 28, 2020 3:56 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: 6N.7b
Replies: 2
Views: 53

Re: 6N.7b

When you write out the half reactions, you will find the anode as H2(g) -> 2H+(aq) +2e- and the cathode as 2H+(aq) + 2e- -> H2(g). When you put them together to find the balanced overall equation, you get 2H+(aq) -> 2H+(aq), where the H2 and e- cancel out from both sides. In this case, you would nee...
by JasonLiu_2J
Fri Feb 28, 2020 3:39 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: 6N 3c)
Replies: 1
Views: 35

Re: 6N 3c)

While we learned to use Q in terms of Q and Qp, you can actually use Q as a combination of concentration and partial pressure as it simply represents a ratio of products to reactants. Note that you want the partial pressure to be in terms of atm/bar (they are essentially equal), so you would have to...
by JasonLiu_2J
Fri Feb 28, 2020 3:34 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: Presence of current
Replies: 1
Views: 19

Re: Presence of current

The electric current produced by a redox reaction is due to the flow of electrons between the two electrolytes in the cell. In a redox reaction, one substance is oxidized while another is reduced. Electrons are thus transferred from the oxidized substance to the substance that is going to be reduced...
by JasonLiu_2J
Fri Feb 28, 2020 3:31 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: Gibbs Free Energy
Replies: 2
Views: 32

Re: Gibbs Free Energy

Just to add on, you can think of the sign of a value as the direction its referring to. A positive value of E indicates the flow of electrons in the forward direction, while a negative value of E indicates the flow of electrons in the reverse direction. When E is negative, electrons flow in the oppo...
by JasonLiu_2J
Fri Feb 28, 2020 3:07 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: finding top of your series
Replies: 3
Views: 27

Re: finding top of your series

In an electrochemical series, the E values are given in terms of the reduction potentials. The strongest oxidizing agent will be the one with the greatest (most positive) reduction potential since it will be the one that is more likely to be reduced. (Remember that the oxidizing agent causes oxidati...
by JasonLiu_2J
Fri Feb 21, 2020 10:40 am
Forum: Balancing Redox Reactions
Topic: HW problem 6k.1
Replies: 3
Views: 22

Re: HW problem 6k.1

In a balanced redox equation that combines the oxidation and reduction half reactions, you want to make sure that there are no leftover electrons, and the electrons in the products of the reduction half reaction and in the reactants of the oxidation half reaction cancel out. Thus, you would need to ...
by JasonLiu_2J
Fri Feb 21, 2020 10:35 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Voltage of 0
Replies: 3
Views: 22

Re: Voltage of 0

The voltage of the cell changes over time as the cell reaction proceeds. When the cell has reached a state of equilibrium, then the reaction is no longer favorable and does not proceed, thus making it so that there is no current in the cell. This is what happens when a battery "dies", as t...
by JasonLiu_2J
Fri Feb 21, 2020 10:31 am
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: 6L.1 (b)
Replies: 1
Views: 21

Re: 6L.1 (b)

I think a large number is expected since Faraday's constant is a very large constant (96,485 C/mol), making the number you get from deltaG=-nFE a very large number.
by JasonLiu_2J
Fri Feb 21, 2020 10:26 am
Forum: Balancing Redox Reactions
Topic: 6k.5 part d
Replies: 2
Views: 42

Re: 6k.5 part d

I believe that when you balanced the H, you added 16H2O instead of 8H2O. Since you already have 8 H atoms on the products side and 16 on the reactant side, you will only need to add 8H2O to the products and 8OH- to the reactants, as opposed to the 16 of each you added initially. This would then get ...
by JasonLiu_2J
Fri Feb 21, 2020 10:17 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Potential Difference
Replies: 2
Views: 23

Re: Potential Difference

The potential difference is the difference in the standard reduction potentials between the two half cells. When the potential difference of the overall cell is positive, you can expect that the overall cell reaction (Transfer of electrons from anode to cathode) is favorable. When the potential diff...
by JasonLiu_2J
Fri Feb 14, 2020 12:30 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 4.17E: Calculating enthalpy change vs. assuming delta U = 0
Replies: 2
Views: 43

Re: 4.17E: Calculating enthalpy change vs. assuming delta U = 0

While it is true that ∆U=0 at constant temperature for an isolated system, enthalpy is only equal to heat under conditions of constant pressure. As such, you can't just say q=-w because you don't know how the pressure changes. You would then have to calculate the enthalpy of the reaction using the s...
by JasonLiu_2J
Fri Feb 14, 2020 12:26 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Differences in Heat Capacity
Replies: 4
Views: 58

Re: Differences in Heat Capacity

It might also be good to know that the heat capacity for a substance is an extensive property, as it depends on how much of the substance is present. Specific heat capacity (energy required to heat a gram by 1 degrees C) and molar heat capacity (energy required to heat a mol by 1 degrees C), are int...
by JasonLiu_2J
Fri Feb 14, 2020 12:15 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Textbook question 4J.17
Replies: 2
Views: 44

Re: Textbook question 4J.17

Whether or not you can calculate the temperature cutoff depends on whether ∆H and ∆S are positive or negative. For the reaction, when ∆H is positive and ∆S is negative, then you know that ∆G must be a positive value, thus making the reaction nonspontaneous no matter the temperature (temperature has ...
by JasonLiu_2J
Fri Feb 14, 2020 12:07 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: reversible expansion vs. irreversible free expansion
Replies: 2
Views: 38

Re: reversible expansion vs. irreversible free expansion

In an isothermal reversible expansion, the system does work on the surroundings when it expands, whereas no work is done on the surroundings in a free expansion. As such, w=0 in a free expansion and the deltaS of the surroundings will also equal zero. However, since deltaS is a state function, the d...
by JasonLiu_2J
Fri Feb 14, 2020 12:00 pm
Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
Topic: reaction entropy
Replies: 7
Views: 76

Re: reaction entropy

Since entropy, like enthalpy, is a state function, you can find entropy of reaction by doing standard entropy of products minus the standard entropy of the reactants (multiplying each by the stoichiometric coefficients. It's also good to know that for enthalpy and Gibbs free energy, you are using th...
by JasonLiu_2J
Fri Feb 07, 2020 10:51 am
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Entropy change at Low Temperature
Replies: 3
Views: 37

Re: Entropy change at Low Temperature

Another way of looking at this is that the same amount of heat (q) will have a greater effect on entropy for a system at low temperature compared to a system at high temperature because the lower temperature system feels the effect of the heat more. A high temperature system has more "heat"...
by JasonLiu_2J
Fri Feb 07, 2020 10:34 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: enthalpy and temperature
Replies: 2
Views: 32

Re: enthalpy and temperature

You probably will not have to predict how enthalpy changes directly from a change in temperature. However, it might be good to know how to calculate the heat change due to a temperature change and relate that to the overall enthalpy change since change in enthalpy=q at constant pressure.
by JasonLiu_2J
Fri Feb 07, 2020 10:20 am
Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
Topic: 4h.11 (d)
Replies: 1
Views: 27

Re: 4h.11 (d)

I think this question has to do with the microstates available to the molecules and the positional entropy. When you compare KClO3 to KCl, you find that KClO3 has many more different positions for the atoms surrounding Cl compared to KCl which can only be in the positions K-Cl or Cl-K. Now comparing...
by JasonLiu_2J
Wed Feb 05, 2020 12:25 pm
Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
Topic: 4.F.17
Replies: 1
Views: 22

Re: 4.F.17

Since you want the entropy of vaporization at 85 degrees celsius, you will need to find the entropy change as the vapor cools back down to 85 degrees after being vaporized at 100 degrees. Without this step, you would have the entropy change when the liquid goes from 85 degrees to 100 degrees and vap...
by JasonLiu_2J
Wed Feb 05, 2020 12:21 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Heat Output
Replies: 1
Views: 21

Re: Heat Output

The heat output of the reaction will be the same as the heat input into the calorimeter, as the heat is transferred from the reaction to the calorimeter. This means that q(reaction) = -q(calorimeter). You can solve for the qcal by using the equation qcal=Ccal*deltaT, which gives you the heat for the...
by JasonLiu_2J
Fri Jan 31, 2020 12:14 pm
Forum: Phase Changes & Related Calculations
Topic: standard reaction enthalpy vs. standard enthalpy of formation
Replies: 7
Views: 69

Re: standard reaction enthalpy vs. standard enthalpy of formation

How will we know when to use standard reaction ethalpy or standard ethalpy of formation? Will the question specifically ask for it? The question will usually specify what it wants you to calculate. Calculating the standard reaction enthalpy can be done in three way (Hess's Law, standard enthalpies ...
by JasonLiu_2J
Fri Jan 31, 2020 12:06 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: Reversible vs Irreversible formula
Replies: 4
Views: 74

Re: Reversible vs Irreversible formula

If the problem indicates that the system is under a constant pressure, then you would use the w=-P∆V equation to calculate the work done. If the problem instead indicates that the reaction is an isothermic expansion where pressure is not constant, then you would use the equation for a reversible rea...
by JasonLiu_2J
Fri Jan 31, 2020 12:00 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: How are qp and qv different?
Replies: 3
Views: 24

Re: How are qp and qv different?

qv can also be used to find the enthalpy under the conditions of constant volume. In this scenario, you would need to use the equation ∆H=∆U + ∆PV. Remember that ∆U=q+w, and at constant volume, w=0 (since w=P∆V). As such, you find that ∆U=qv, and the equation for enthalpy change becomes ∆H=qv+V∆P, w...
by JasonLiu_2J
Fri Jan 31, 2020 11:50 am
Forum: Calculating Work of Expansion
Topic: w=-P∆V
Replies: 2
Views: 30

Re: w=-P∆V

W= -P∆V is used for reactions where the external pressure is constant. This is the case for an irreversible process that is under a constant external pressure. A reversible isothermic expansion is not at a constant pressure, so you must instead use the other equation for work which is w=-nRTln(V2/V1)
by JasonLiu_2J
Fri Jan 31, 2020 11:34 am
Forum: Administrative Questions and Class Announcements
Topic: Midterm
Replies: 8
Views: 70

Re: Midterm

Dr. Lavelle just made an announcement on his website that the midterm will cover Equilibrium, Acids and Bases, Thermochemistry, and Thermodynamics.
by JasonLiu_2J
Thu Jan 30, 2020 3:21 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 4C.3 Change in Enthalpy
Replies: 3
Views: 92

Re: 4C.3 Change in Enthalpy

To find the change in enthalpy at constant volume, you would need to use the equation ∆ H = ∆ U +nR∆T. ∆U = q+w but since we know that volume is constant, we know that the work done will be zero (since w=-P∆V). Thus, ∆U=q. You would then plug in the given values of q, n, and R as well as your calcul...
by JasonLiu_2J
Tue Jan 21, 2020 11:50 am
Forum: Acidity & Basicity Constants and The Conjugate Seesaw
Topic: Ka*Kb=Kw
Replies: 5
Views: 58

Re: Ka*Kb=Kw

It does not make a difference for calculations since 10^-14 and 1.0 x 10^-14 are the same thing but you should note that 1.0 x 10^-14 gives you two sig figs which would be useful to have the correct number of sig figs in your answer.
by JasonLiu_2J
Tue Jan 21, 2020 11:48 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: How to find percentage ionization of a basic solution
Replies: 2
Views: 20

Re: How to find percentage ionization of a basic solution

Percentage deprotonation only applies to acids since acids donate a proton in solution. A base does not donate protons, but rather accepts protons, so you can only find percentage protonation for bases. Thus, percentage deprotonation is for acids whereas percentage protonation is for bases. However,...
by JasonLiu_2J
Tue Jan 21, 2020 11:41 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 6A.21
Replies: 2
Views: 20

Re: 6A.21

Kw is equal to the concentration of OH- ions multiplied by the concentration of H3O+ ions. While Kw varies with temperature, the concentrations of the two ions will still be equal to each other for water. Therefore, you would simply take the square root of Kw to find the concentrations of the two io...
by JasonLiu_2J
Mon Jan 20, 2020 10:50 am
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Le Chatelier's Principle
Replies: 2
Views: 18

Re: Le Chatelier's Principle

That is correct. Changing pressure only causes an equilibrium shift when the change in pressure comes from a change in volume, since this also leads to a change in concentration for the reactants and products. Changing pressure through the addition of an inert gas which doesn't affect concentrations...
by JasonLiu_2J
Mon Jan 20, 2020 10:41 am
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: 5J.1
Replies: 2
Views: 26

Re: 5J.1

Remember that the partial pressure of a substance is directly proportional to the moles of that substance through the equation PV=nRT, or P=nRT/V. As such, increasing the partial pressure of a substance is really increasing the concentration of that substance. When you increase the partial pressure ...
by JasonLiu_2J
Thu Jan 16, 2020 8:51 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: ice table/quadratic
Replies: 6
Views: 50

Re: ice table/quadratic

It is definitely possible for both values of x from the quadratic equation to be positive. In that scenario, you have to check to see whether or not the values of x are larger than the initial concentration. If the value of x is larger than the initial concentration, then that value cannot be used b...
by JasonLiu_2J
Thu Jan 16, 2020 8:46 pm
Forum: Ideal Gases
Topic: partial pressure
Replies: 1
Views: 28

Re: partial pressure

The partial pressure of a gas is the pressure that the gas would exert if it were to occupy the total volume alone. That is, it is simply the amount of pressure that gas contributes to the total pressure in the container, and is not affected by the other gases/molecules present. If you increased the...
by JasonLiu_2J
Thu Jan 16, 2020 4:32 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Q and K
Replies: 4
Views: 27

Re: Q and K

When Q<K, it means there are more reactants and less products compared to equilibrium, so the reaction will proceed to the right and favor product formation. When Q>K, there are more products and less reactants compared to the equilibrium concentrations, so the reaction will proceed to the left and ...
by JasonLiu_2J
Wed Jan 15, 2020 11:22 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 5.35 finding partial pressure
Replies: 2
Views: 17

Re: 5.35 finding partial pressure

The y-axis on the graph gives the partial pressures in terms of kPa. Since 1pKA is equal to 0.01 bar, you can convert the equilibrium partial pressures in the graph from pKa to bars and use those partial pressures for the equilibrium constant equation.
by JasonLiu_2J
Wed Jan 15, 2020 11:18 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Ice Box
Replies: 9
Views: 70

Re: Ice Box

The given concentrations won't always have a negative change in concentration since sometimes you are given an initial concentration of both products and reactants and asked to calculate the equilibrium concentrations. In such a case, you would need to calculate Q using the initial concentrations of...
by JasonLiu_2J
Thu Jan 09, 2020 2:25 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Textbook Self-test 5G.3A
Replies: 2
Views: 29

Re: Textbook Self-test 5G.3A

Just to add on, since the Ag2O is in solid phase rather than aqueous, it does not dissociate into its ions in an aqueous solution. It would still be considered Ag2O when writing the net ionic equation, since an ion is only counted as a spectator if the dissociated ion itself is present on both sides...
by JasonLiu_2J
Wed Jan 08, 2020 11:19 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 5G.1
Replies: 1
Views: 21

Re: 5G.1

The equilibrium constant is a set value that is not affected by the initial amount of reactant or product. No matter the initial concentration or partial pressure, the equilibrium concentrations or partial pressures will have the same ratio once the reaction has reached equilibrium (as long as tempe...
by JasonLiu_2J
Wed Jan 08, 2020 9:43 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 5I.13
Replies: 2
Views: 28

Re: 5I.13

When 2.0mmol of F2 is added to the reaction vessel with 2.0L, you find that the initial concentration of F2 is indeed 0.001 M. However, this does not give the equilibrium concentration, which is what you want to find. In this case, you would need to set up the dissociation equation of F2<->2F and us...
by JasonLiu_2J
Wed Jan 08, 2020 9:36 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: 5I.9
Replies: 3
Views: 37

Re: 5I.9

I think that since the problem gives the reactants and products in terms of partial pressures, the given K value will also refer to Kp. Generally, the K value should correspond to the units given in the problem unless it is specifically stated in the question that K is either Kp or Kc.
by JasonLiu_2J
Wed Jan 08, 2020 9:32 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 5.35
Replies: 4
Views: 38

Re: 5.35

From the question you know that compound A decomposes into compounds B and C. This indicates that A is the reactant whereas B and C are the products of the reaction. If you compare how much the partial pressure of A decreases compared with how much the partial pressures of B and C increase, you can ...
by JasonLiu_2J
Thu Dec 05, 2019 1:28 pm
Forum: Hybridization
Topic: Hybridization in Pi Bonds
Replies: 4
Views: 41

Re: Hybridization in Pi Bonds

Yea there are 3 hybridized sp2 orbitals. Sorry for the typo
by JasonLiu_2J
Tue Dec 03, 2019 6:09 pm
Forum: Hybridization
Topic: Hybridization of double & triple bonds
Replies: 1
Views: 13

Re: Hybridization of double & triple bonds

Yes for triple bonds you would write the pi(atom 2p, atom 2p) label twice to indicate both pi bonds.
by JasonLiu_2J
Mon Dec 02, 2019 3:58 pm
Forum: Identifying Acidic & Basic Salts
Topic: Knowing whether a salt is acidic or basic
Replies: 1
Views: 27

Re: Knowing whether a salt is acidic or basic

The ability of the salt to affect the pH of a solution does to an extent depend on whether the salt consists of strong conjugate acids/bases. Strong conjugate acids/bases result from weak acids and weak bases, which means that these conjugates will have a greater ability to affect water molecules an...
by JasonLiu_2J
Mon Dec 02, 2019 3:50 pm
Forum: Lewis Acids & Bases
Topic: charged acid
Replies: 1
Views: 23

Re: charged acid

Whether the acid is neutral or negatively charged depends on if it is protonated or not. For example, HCl is a strong acid. In this form, it is considered neutral because it does not have a positive or negative charge associated with it. However, when HCl is deprotonated and separated into H+ and Cl...
by JasonLiu_2J
Mon Dec 02, 2019 3:12 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Subshell Notation
Replies: 2
Views: 57

Re: Subshell Notation

I don't think you can find the hybridization of an atom based purely on its quantum numbers since hybridization depends on the bonding that the atom exhibits with other atoms. For example, the valence electrons of the carbon atom will have quantum numbers n=2, l=1, ml= -1, 0, +1, and ms= +/- 1/2. Ho...
by JasonLiu_2J
Mon Dec 02, 2019 2:59 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: H3O versus OH
Replies: 1
Views: 28

Re: H3O versus OH

You can calculate the concentration of OH- when given [H3O+] by using the equation Kw=[OH-]*[H3O+]. Kw is equal to 1.0 x 10^-14. By dividing this value by the concentration of H3O+, you will have the concentration of OH- in solution. Converting to pH, pOH, then finally [OH-] would also work but it t...
by JasonLiu_2J
Mon Dec 02, 2019 2:55 pm
Forum: Properties & Structures of Inorganic & Organic Acids
Topic: Acid strength
Replies: 2
Views: 37

Re: Acid strength

The strength of the acid depends on how easy it is to deprotonate, or remove the hydrogen atom, from the acid. In the case of HF vs HCl, the strength of the bond is determined by the size of the atoms since they are inorganic molecules and the acid anions are in the same group on the periodic table....
by JasonLiu_2J
Wed Nov 27, 2019 10:55 am
Forum: Hybridization
Topic: Hybridization in Pi Bonds
Replies: 4
Views: 41

Re: Hybridization in Pi Bonds

A pi bond is formed when two p-orbitals overlap side to side. When C double bonds with C as in ethylene, the C atom has sp2 hybridization, leaving one unhybridized p-orbital with 3 hybridized sp orbitals. These sp orbitals form sigma bonds with other atoms while the unhybridized p-orbital of one C a...
by JasonLiu_2J
Wed Nov 27, 2019 10:46 am
Forum: Dipole Moments
Topic: H-bonds and H-bonding
Replies: 2
Views: 67

Re: H-bonds and H-bonding

Hydrogen bonding is not a concrete bond between two atoms, but rather an extremely strong attractive force between a hydrogen atom that is covalently bonded to N, O, or F and the lone pairs on another N, O, or F atom on another molecule. The N, O, and F are very electronegative, which gives the H at...
by JasonLiu_2J
Wed Nov 27, 2019 10:41 am
Forum: Properties & Structures of Inorganic & Organic Bases
Topic: Nonmental Oxides as Acids
Replies: 4
Views: 47

Re: Nonmental Oxides as Acids

Yes, CO2 would be considered an acid by the Lewis acid definition. In the reaction, the central carbon atom on CO2 accepts a lone pair from the O in H2O, making the CO2 molecule an electron pair acceptor, or Lewis Acid. At the same time, one of the Hydrogen atoms is delocalized and forms a bond with...
by JasonLiu_2J
Wed Nov 27, 2019 10:25 am
Forum: Bronsted Acids & Bases
Topic: Forming a neutralization reaction
Replies: 2
Views: 41

Re: Forming a neutralization reaction

It would be good to have an understanding of how to write the neutralization reaction in case it does show up on the test since it is very applicable to acids and bases and was part of homework set problems.
by JasonLiu_2J
Mon Nov 25, 2019 3:24 pm
Forum: Properties & Structures of Inorganic & Organic Acids
Topic: How do you know is an acid is strong or weak?
Replies: 6
Views: 66

Re: How do you know is an acid is strong or weak?

Strong acids almost completely dissociate in water whereas weak acids only partially dissociate. There are really only a certain number of strong acids. These are HBr, HCl, HI, HNO3, HClO4, HClO3, H2SO4. Any other acids would be considered weak acids.
by JasonLiu_2J
Fri Nov 22, 2019 10:04 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: LDF vs dipole-dipole
Replies: 4
Views: 43

Re: LDF vs dipole-dipole

Yes, the strength of the LDF forces can have a greater influence on the relative boiling points of two molecules than the difference in dipole dipole forces. Consider the example of HCl vs HI. Since Cl is more electronegative than I, the HCl molecule would be more polar and thus, have a stronger dip...
by JasonLiu_2J
Fri Nov 22, 2019 9:44 am
Forum: *Molecular Orbital Theory (Bond Order, Diamagnetism, Paramagnetism)
Topic: sigma bond
Replies: 7
Views: 99

Re: sigma bond

Sigma bonds occurs when two orbitals interact/overlap end-to-end, which typically leads to them having one region of shared electron density. Imagine a venn diagram where the two circles are the orbitals of two atoms and the overlapping region is the region of shared electron density. This middle sh...
by JasonLiu_2J
Fri Nov 22, 2019 9:16 am
Forum: Determining Molecular Shape (VSEPR)
Topic: 2E.5
Replies: 5
Views: 39

Re: 2E.5

Determining the VSEPR formula is pretty helpful in finding the molecular shape. For ClO2+, you know from the lewis structure that it is bonded to two oxygens with one lone pair on the central Cl atom. This indicates that it would have a VSEPR formula of AX2E, which corresponds to a bent structure. A...
by JasonLiu_2J
Tue Nov 19, 2019 7:07 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Axial vs equatorial lone pairs
Replies: 1
Views: 34

Re: Axial vs equatorial lone pairs

The axial and equatorial planes are important when considering the trigonal bipyramidal shape. The two atoms at the top and bottom of the shape that lie 180 degrees from each other are on the axial plane (think the axis that the world spins on) whereas the three atoms in the middle 90 degrees from t...
by JasonLiu_2J
Tue Nov 19, 2019 5:54 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Nitric Oxide
Replies: 1
Views: 19

Re: Nitric Oxide

Since a nitric oxide molecule only has two atoms, it must have a linear structure. The molecule itself is polar because of the electronegativity difference between nitrogen and oxygen, which gives the molecule a dipole moment pointing towards oxygen, the more electronegative atom. The fact that this...
by JasonLiu_2J
Thu Nov 14, 2019 3:16 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: format of test
Replies: 5
Views: 54

Re: format of test

It should be about the same length of test 1 and will most likely cover the information we have learned after the midterm. This would include information such as IMFs and VSEPR structures, as well as whatever we learn in class this week before the test
by JasonLiu_2J
Thu Nov 14, 2019 3:12 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Bond Angle
Replies: 3
Views: 30

Re: Bond Angle

This is a very broad question since there are many, many VSEPR shapes depending on different combinations of numbers of bonding pairs and lone pairs on the central atom. However, for the basic structures with no lone pairs, they are as follows: 2 atoms attached to central atom = linear = 180 degrees...
by JasonLiu_2J
Thu Nov 14, 2019 3:03 pm
Forum: Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)
Topic: How to remember strength of different intermolecular forces
Replies: 5
Views: 44

Re: How to remember strength of different intermolecular forces

Another way to remember the order of intermolecular force strength is to think about the strength of the charges on ions, dipoles, and induced dipoles respectively. Ions are charged atoms of elements, and thus, already have a full charge associated with them (+1, +3, -2 etc.) Thus, the interactions ...
by JasonLiu_2J
Wed Nov 13, 2019 5:20 pm
Forum: Bond Lengths & Energies
Topic: Boiling Points (HW 3F.5)
Replies: 2
Views: 30

Re: Boiling Points (HW 3F.5)

The difference between HCF3 and HCI3 is that one has fluorine atoms and the other has iodine atoms. Iodine is larger than fluorine, which makes it more polarizable and increases the strength of the induced dipole induced dipole attraction. Stronger attractions leads to a higher melting point since i...
by JasonLiu_2J
Wed Nov 13, 2019 4:53 pm
Forum: Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)
Topic: 3F Question 1c
Replies: 1
Views: 30

Re: 3F Question 1c

H2SeO4 also exhibits hydrogen bonding. This is because when you draw the lewis structure for H2SeO4, you will have the Se double bonded with two O atoms, and single bonded with the other 2 O atoms. The two hydrogens are then have a single bond with the 2 single-bonded oxygens. This lewis structure g...
by JasonLiu_2J
Tue Nov 05, 2019 9:05 pm
Forum: Quantum Numbers and The H-Atom
Topic: Quantum numbers
Replies: 1
Views: 34

Re: Quantum numbers

The first two quantum numbers are the n and l quantum numbers. These indicate the energy level and subshell in which the electron resides. For the last electron in the element, you are looking at the valence electron. Thus it would be the farthest orbital, or rather, the first electron that you woul...
by JasonLiu_2J
Tue Nov 05, 2019 4:45 pm
Forum: Ionic & Covalent Bonds
Topic: 2D.9
Replies: 3
Views: 39

Re: 2D.9

An ion has higher polarizing power when it has a higher charge as well as when is smaller. Smaller ions (those with smaller ionic radii) have a higher polarizing power because the nucleus is closer to the electrons of the anion, thus causing a greater attraction of the electrons towards the cation. ...
by JasonLiu_2J
Tue Nov 05, 2019 2:50 pm
Forum: Quantum Numbers and The H-Atom
Topic: Hydrogen Atom absorbing a photon
Replies: 1
Views: 25

Re: Hydrogen Atom absorbing a photon

I think the energy of the photon must be equal to the difference in energy levels for the electron to be excited to a different energy level. If the energy of the photon is between two energy levels, the electron would not be excited to a higher energy level since electrons must exist in discrete en...
by JasonLiu_2J
Mon Nov 04, 2019 3:01 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Heisenberg Uncertainty Quantum World Worksheet
Replies: 3
Views: 80

Re: Heisenberg Uncertainty Quantum World Worksheet

Whether or not you can blame Heisenberg's Uncertainty Principle depends on the relative scales of your calculated indeterminacy and the values of the position/momentum for the object you are dealing with. For something as large as a bowling ball, an indeterminacy value on the scale of, for example, ...
by JasonLiu_2J
Mon Nov 04, 2019 8:57 am
Forum: Ionic & Covalent Bonds
Topic: Valence Electrons and Electron Configuration
Replies: 1
Views: 27

Re: Valence Electrons and Electron Configuration

I believe that when counting valence electrons in an atom, you only consider those in the outermost shell/energy level. In Ga for example, since it only has 3 electrons in its 4s and 4p orbitals, it would only have 3 valence electrons. The 3d electrons are part of the n=3 energy level, and would not...
by JasonLiu_2J
Mon Nov 04, 2019 8:48 am
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Uncertainty Principle
Replies: 1
Views: 40

Re: Uncertainty Principle

That's correct. When given a +/- value, the uncertainty is the range of possible values, so you can simply calculate it by doing the +/- value multiplied by two. This is also true when you have a +/- when given uncertainty in position.
by JasonLiu_2J
Sat Nov 02, 2019 3:36 pm
Forum: Electronegativity
Topic: Oxygen and Electronegativity
Replies: 5
Views: 37

Re: Oxygen and Electronegativity

The trend for electronegativity is that it increases up a group and from left to right across a period. Since oxygen is towards the upper right of the periodic table, it has a high electronegativity. This can also be explained conceptually when you consider an oxygen atom. An oxygen atom is small, w...
by JasonLiu_2J
Thu Oct 31, 2019 1:28 pm
Forum: Properties of Light
Topic: HW #1.31
Replies: 1
Views: 42

Re: HW #1.31

You are correct in using the 405nm wavelength light since it has higher energy and is able to eject electrons from the lithium surface. One thing to consider when dealing with this problem is the units you use. The energy of the light you use, hv, will inevitably be small because it indicates the en...
by JasonLiu_2J
Wed Oct 30, 2019 5:11 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: 1E.25 Part C
Replies: 2
Views: 32

Re: 1E.25 Part C

I think what they mean by group 5 transition metals is specifically the fifth group of the transition metals, not the fifth group of the periodic table. If you take it as being the fifth group in the d-block, then you will get (n-1)d^5 ns^2.
by JasonLiu_2J
Wed Oct 30, 2019 2:27 pm
Forum: Dipole Moments
Topic: Boiling Points
Replies: 1
Views: 8

Re: Boiling Points

As you go down a group, the atoms increase in size, which causes the London forces to also increase in strength. This is because the electrons are farther away from the nucleus and less tightly held to the atom, allowing there to be a much larger area for dipole moments and making it easier for larg...
by JasonLiu_2J
Tue Oct 29, 2019 11:47 pm
Forum: Formal Charge and Oxidation Numbers
Topic: formal charge
Replies: 5
Views: 65

Re: formal charge

Another way to determine formal charge is through a variation of the equation FC=V-(L+S/2). One way to think about it is to use number of bonds instead of S/2. The way that I usually calculate formal charge is by doing FC=V-(L+ #of bonds). S/2 is the same as simply counting each bond as 1 (since you...
by JasonLiu_2J
Tue Oct 29, 2019 11:41 pm
Forum: Lewis Structures
Topic: 2B.3d
Replies: 1
Views: 25

Re: 2B.3d

One of the exceptions to the octet rule that can occur is through expanded valence shells. Elements that have electrons in energy level 3 (n=3)and beyond can make use of d-orbitals to hold valence shell electrons, since n=3 corresponds to l=2, and thus, d-orbitals. Since Br is in energy level 4, it ...
by JasonLiu_2J
Thu Oct 24, 2019 3:42 pm
Forum: Lewis Structures
Topic: Radicals and identifying them
Replies: 1
Views: 39

Re: Radicals and identifying them

Generally speaking, a radical occurs when the total number of valence electrons in the molecule adds up to an odd number, as this provides no way in which you can complete the octet rule. For NO2-, N has 5 valence electrons, O2 has 12 in total, and the minus charge adds another electron, giving you ...
by JasonLiu_2J
Tue Oct 22, 2019 5:18 pm
Forum: Photoelectric Effect
Topic: module assessment Q #29
Replies: 2
Views: 38

Re: module assessment Q #29

The work function for sodium is given in terms of kJ/mol. However, in order to calculate the energy needed to eject an electron from one sodium atom, you have to convert from kJ/mol to J/electron. Remember that with the photoelectric effect, you can think of the relationship between photons and elec...
by JasonLiu_2J
Tue Oct 22, 2019 5:11 pm
Forum: Trends in The Periodic Table
Topic: Hw Help 1F.22
Replies: 2
Views: 23

Re: Hw Help 1F.22

You can tell whether most elements are metals, metalloids, or nonmetals based on their properties which you can determine through their placement on the periodic table. Metals have a low ionization energy, which allows them to be highly reactive and to form cations (lose electrons). These are genera...
by JasonLiu_2J
Tue Oct 22, 2019 4:40 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: HW 1E.5
Replies: 2
Views: 33

Re: HW 1E.5

To add on to what was previously said, electrons in the s-orbital has a symmetrical electron density distribution in a spherical shape about the nucleus. That is, the s-orbital electrons have a greater ability to penetrate to the nucleus because they have a higher probability to be found at or near ...
by JasonLiu_2J
Mon Oct 21, 2019 11:17 am
Forum: DeBroglie Equation
Topic: 1B. 19
Replies: 4
Views: 50

Re: 1B. 19

That is correct. According to the De Broglie Wavelength equation, λ=h/p, where p, aka the momentum, is equal to mass x velocity. Since h is a constant, the only factor that affects the De Broglie wavelength of a particle would be its mass and velocity. If two particles have the same mass and velocit...
by JasonLiu_2J
Mon Oct 21, 2019 10:15 am
Forum: *Shrodinger Equation
Topic: Hamiltonian
Replies: 6
Views: 54

Re: Hamiltonian

As far as I know, we won't need to do any calculations using Schrodinger's equation or the Hamiltonian in this class.
by JasonLiu_2J
Tue Oct 15, 2019 1:52 pm
Forum: Photoelectric Effect
Topic: Intensity, frequency, and wavelength
Replies: 5
Views: 59

Re: Intensity, frequency, and wavelength

When you view light as particles, the factor that influences the energy of the photon is the frequency of the light used. You can see this relationship from the equation E=hv, in which energy and frequency are directly proportional. Thus, as you increase the frequency, the energy of the photon also ...
by JasonLiu_2J
Tue Oct 15, 2019 1:45 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: What does "l" mean?
Replies: 2
Views: 61

Re: What does "l" mean?

The value of "l" corresponds to the orbital angular momentum. It essentially tells you in which subshell the electron resides, whether it be the s-, p-, d-, or f- subshells. When l=0, the electron is part of the s-subshell. If l=1, then the electron is part of the p-subshell and so on. Eac...
by JasonLiu_2J
Tue Oct 15, 2019 1:37 pm
Forum: Photoelectric Effect
Topic: Question on Example 1B.3
Replies: 1
Views: 35

Re: Question on Example 1B.3

Remember that wavelength and frequency are inversely proportional (from c=λv). Since part c asks for the longest wavelength of electromagnetic radiation that could eject electrons from potassium, you could also interpret this as meaning what is the smallest frequency that could eject the electrons. ...
by JasonLiu_2J
Mon Oct 14, 2019 3:17 pm
Forum: Photoelectric Effect
Topic: 1B.15
Replies: 2
Views: 42

Re: 1B.15

Just to add on, to find the wavelength of the ejected electron, you can also use the De Broglie equation, which provides a relationship between wavelength and momentum. The De Broglie equation indicates that the wavelength is equal to Planck's constant divided by the momentum of the object in questi...
by JasonLiu_2J
Mon Oct 14, 2019 10:18 am
Forum: Quantum Numbers and The H-Atom
Topic: Question 1D.17
Replies: 1
Views: 37

Re: Question 1D.17

The possible values for the magnetic quantum number is based on the value of the orbital angular momentum "l". The possible values include the positive and negative values of l and all integers in between the two. For example, for the orbital 6p, the p orbital corresponds to an orbital ang...
by JasonLiu_2J
Sun Oct 13, 2019 12:28 pm
Forum: Photoelectric Effect
Topic: Intensity, frequency, and wavelength
Replies: 5
Views: 59

Re: Intensity, frequency, and wavelength

In the photoelectric experiment, light is acting as a particle in the form of photons. In order for an electron to be emitted, each individual photon must have sufficient energy that is equal to or greater than that required to eject the electron from the surface. If the photons do not have enough e...
by JasonLiu_2J
Tue Oct 08, 2019 2:02 pm
Forum: Molarity, Solutions, Dilutions
Topic: Question on online problems for Dilution
Replies: 2
Views: 83

Re: Question on online problems for Dilution

Remember that the units for molarity are moles/liter (number of moles/volume of solution). From the question, you know that you have 800ml, or 0.8 L of solution. What you need to find is how many moles of ethanol are in 12.4 grams of ethanol (CH3CH2OH). To do this, you would divide the amount in gra...
by JasonLiu_2J
Tue Oct 08, 2019 1:56 pm
Forum: Limiting Reactant Calculations
Topic: Limiting Reactant Problem
Replies: 5
Views: 67

Re: Limiting Reactant Problem

The first thing you want to do in this question after balancing the equation is to find the limiting reactant based on your starting amount of each reactant. Convert the amount in grams of each reactant to the number of moles of each reactant that you initially start with. This is done by dividing t...

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