Search found 45 matches
- Sun Feb 14, 2021 9:57 pm
- Forum: Van't Hoff Equation
- Topic: Equation Sheet
- Replies: 18
- Views: 1122
Re: Equation Sheet
No, it's not on the equation sheet, but we do have to know it so either know how to derive it or just memorize it.
- Sun Feb 14, 2021 9:53 pm
- Forum: Administrative Questions and Class Announcements
- Topic: struggling
- Replies: 73
- Views: 3942
Re: struggling
I'm stressed and struggling too, so you're definitely not alone. I recommend attending UA sessions as they're super helpful, as well as checking your confidence/understanding of concepts. Remember, grades don't define you, so don't stress too much and remember to pace yourself and not be too hard on...
- Sun Feb 14, 2021 9:48 pm
- Forum: Administrative Questions and Class Announcements
- Topic: cc update
- Replies: 23
- Views: 1223
Re: cc update
They'll probably be updated soon, so don't stress too much about them not being accurate at the moment.
- Sun Feb 14, 2021 9:44 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Midterm 2
- Replies: 33
- Views: 2216
Re: Midterm 2
Any good advice or tips for studying this week? I'm trying not to stress too much.
- Sun Feb 14, 2021 9:40 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Likeliness to form products/reactants
- Replies: 27
- Views: 966
Re: Likeliness to form products/reactants
When delta G is negative, the forward reaction and products are favored, and when delta G is positive, the reactants and reverse reaction is favored.
- Sun Feb 14, 2021 9:37 pm
- Forum: Ideal Gases
- Topic: Chem BL
- Replies: 107
- Views: 8277
Re: Chem BL
I'm hoping to take it either next quarter or over the summer.
- Sun Feb 14, 2021 9:36 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Including n
- Replies: 11
- Views: 546
Re: Including n
We mostly never ignore n, as it stands for moles. Also, pay attention to the units the question is asking for.
- Sun Feb 14, 2021 9:32 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Is it necessary to memorize Cp,m or Cv,m values? [ENDORSED]
- Replies: 26
- Views: 1340
Re: Is it necessary to memorize Cp,m or Cv,m values? [ENDORSED]
The values are given to us on exams, so most likely you don't need to memorize them.
- Sun Feb 14, 2021 9:31 pm
- Forum: Student Social/Study Group
- Topic: Fav Shows of the Moment
- Replies: 115
- Views: 47963
Re: Fav Shows of the Moment
Hospital Playlist is great! I am also hoping to rewatch Reply 1988 one of these days.
- Sun Feb 14, 2021 9:26 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: G° vs G
- Replies: 30
- Views: 2669
Re: G° vs G
Gº is under standard conditions, whereas G isn't.
- Sun Feb 14, 2021 9:21 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Can We Review our Exams?
- Replies: 69
- Views: 3377
Re: Can We Review our Exams?
Meet with your TA during their office hours!
- Sun Feb 14, 2021 9:18 pm
- Forum: Student Social/Study Group
- Topic: Silly Mistakes?
- Replies: 72
- Views: 6263
Re: Silly Mistakes?
Thank you so much for posting this. It really helped me examine my state of mind and how hard on myself I sometimes get. Exams are always stressful for me, so this provided some insight on my worries.
- Sun Feb 14, 2021 9:13 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: temperature
- Replies: 32
- Views: 1354
Re: temperature
This mostly depends on what units the question is asking for, but most of the time, convert to Kelvins.
- Sun Jan 24, 2021 11:15 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Neutral Solution
- Replies: 9
- Views: 795
Re: Neutral Solution
A solution is neutral when the concentration of H3O+ is 1.0x10^-7, which is important because when [H3O+] = 1.0x10^-7, the pH is 7, and [OH-] = 1.0x10^-7, and pOH = 7. Kw is the concentration constant of neutral water at 25º C, which is 1.0x10^-14, which can only be true when concentrations of [H3O+...
- Sun Jan 24, 2021 11:08 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Percent Ionization
- Replies: 7
- Views: 371
Re: Percent Ionization
Percent ionization is found by dividing the concentration of [H+] at equilibrium by the initial concentration of the acid and multiplying by 100 (to get it in %).
- Sun Jan 24, 2021 10:50 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Fridays Lecture Clarification
- Replies: 3
- Views: 246
Re: Fridays Lecture Clarification
I agree with what John said. I believe Dr. Lavelle referred to standard states to explain how when something is not in its standard state, one needs to take into account the phase change as well when calculating change in enthalpy.
- Sun Jan 24, 2021 10:43 pm
- Forum: Student Social/Study Group
- Topic: Midterm Study Tips
- Replies: 41
- Views: 1754
Re: Midterm Study Tips
Definitely attending UA sessions and Step-up sessions, as well as practicing textbook problems to just solidify your understanding of the topics covered. I personally go over problems that I have difficulty with and practicing with similar problems in the textbook until I feel comfortable with the c...
- Sun Jan 24, 2021 10:36 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Sapling Homework Week 2
- Replies: 5
- Views: 349
Re: Sapling Homework Week 2
You can generally assume x is negligible if the K value is 1000x (10^3) smaller than the initial concentration given, or if the K value is extremely small, like the example Lavelle showed in class where the K value was 10^-30-something. However, it's always a good idea to check if x is really neglig...
- Sun Jan 24, 2021 10:23 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Material Accessible during Exams
- Replies: 6
- Views: 303
Re: Material Accessible during Exams
I think Ka and Kb values would be given in the problems. I don't believe we need to memorize any specific values. However, it would be good to know strong acids and bases, as well as what a larger or smaller Ka value relates to how acidic a compound is.
- Sun Jan 17, 2021 8:36 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Sapling Week 1 #9
- Replies: 4
- Views: 447
Re: Sapling Week 1 #9
So first, find K and set up an ICE table. K is found from [NO]^2/[N2][O2], so it would be [0.400]^2/[0.300][0.300] = 16/9 Products [NO] are added, so reaction shifts toward reactants. Setting up an ICE table you get: [N2]=[O2]=(0.300 + x) and [NO]=(0.700-2x) *keep in mind that change in NO is -2x be...
- Sun Jan 17, 2021 8:20 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Sapling Week 1 #9
- Replies: 11
- Views: 791
Re: Sapling Week 1 #9
The change in x would be +x for both N2 and O2, while for 2NO the change in x would be -2x. This is because adding 2NO to the reaction at equilibrium causes the reaction to shift towards the reactants according to Le Chatelier's Principle. The molar coefficients for N2 and O2 are both 1 so the chang...
- Sun Jan 17, 2021 8:03 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: ICE table equilibrium
- Replies: 8
- Views: 372
Re: ICE table equilibrium
Basically, you want to look at what exactly is changing/being added to the reaction in order to figure out the change in molarity. The change will not always be positive for products and negative for reactants. Looking at a reaction one would generally ask, does the reaction move towards making more...
- Sun Jan 17, 2021 7:55 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Q and K
- Replies: 46
- Views: 1767
Re: Q and K
Yes, Q is the reaction quotient, and K is the equilibrium constant. Q indicates that there is a shift, while K indicates equilibrium. We use Q and compare it to K to see whether the reaction is at equilibrium (Q=K), favors reactants (Q>K), or favors products (Q<K).
- Sun Jan 17, 2021 7:51 pm
- Forum: Student Social/Study Group
- Topic: Comfort Movies
- Replies: 168
- Views: 27598
Re: Comfort Movies
I definitely love watching Pride and Prejudice, Your Name, and Howl's Moving Castle. I love romances and any Ghibli films are always a good watch for feelings of nostalgia and comfort.
- Sun Jan 17, 2021 7:47 pm
- Forum: Ideal Gases
- Topic: reversing reactions
- Replies: 83
- Views: 5332
Re: reversing reactions
It would be the inverse of K, so 1/K.
- Sun Jan 17, 2021 7:46 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Sapling Homework #5
- Replies: 9
- Views: 380
Re: Sapling Homework #5
When combining equations, keep in mind that the K value of a reverse reaction will be the inverse of the original K value: (1/K), while the K value of a reaction that has a coefficient (c) will be the original reaction's K value raised to the power of the coefficient: K^c. Then, once the new K value...
- Sun Jan 10, 2021 11:46 pm
- Forum: Ideal Gases
- Topic: Reading the textbook
- Replies: 262
- Views: 150732
Re: Reading the textbook
The textbook can definitely be helpful supplemental material to support learning. I generally us it to reaffirm knowledge of concepts and getting more practice and comfortable with problems.
- Sun Jan 10, 2021 11:40 pm
- Forum: Student Social/Study Group
- Topic: Study Tips
- Replies: 32
- Views: 1241
Re: Study Tips
Definitely doing practice problems help, as well as making sure one understands a concept and is comfortable enough to solve problems before moving on.
- Sun Jan 10, 2021 11:35 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: K vs. Q
- Replies: 53
- Views: 2148
Re: K vs. Q
Devan Nathu - 2H wrote:Just to clarify, will we always be able to assume that the reaction is at equilibrium when K=Q? Thanks!!
Yes, since K is the equilibrium constant, when Q (the reaction quotient) is equal to K, the reaction is at equilibrium.
- Sun Jan 10, 2021 11:24 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 3612416
Re: Post All Chemistry Jokes Here
I really enjoy chemistry jokes, periodically, that is.
- Sun Jan 10, 2021 11:06 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Advice from a Medical Student - Part II [ENDORSED]
- Replies: 298
- Views: 265356
Re: Advice from a Medical Student - Part II [ENDORSED]
This is really inspiring to read about, and I am so grateful to be able to read your advice and experiences.
- Sat Dec 07, 2019 11:38 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Linear VSEPR model
- Replies: 21
- Views: 1349
Re: Linear VSEPR model
So if the central atom has lone pairs in addition to the bods, then the molecular shape will be bent due to the additional repulsion that the lone pairs have on the other atoms/bonded pair electrons.
- Sat Dec 07, 2019 11:34 pm
- Forum: Naming
- Topic: charges on a compound
- Replies: 2
- Views: 184
Re: charges on a compound
So you’d find the charge on the cation by calculating the charge of the anion and figuring out what the charge should be on the cation to get the total charge of the ligand. In this case the total charge is 2-, and CN has a -1 charge (and there are 5 CN- anions, adding up to a -5 charge). Therefore,...
- Sat Dec 07, 2019 11:24 pm
- Forum: Naming
- Topic: bis,tris,tetrakis
- Replies: 6
- Views: 511
Re: bis,tris,tetrakis
They are prefixes used for naming ligand when the name for a ligand already includes di- (for example, one wouldn’t say di-di-ethylene, but rather bis-diethylene), tri-, or tetra-, and also when a ligand is polydentate (like en/ethylenediamine).
- Sat Dec 07, 2019 11:20 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: bond angles
- Replies: 16
- Views: 1216
Re: bond angles
Bond angles were covered in lecture and can also be found in the textbook. Bond angles are affected by repulsion between electrons, with stronger repulsions between lone pair and lone pair electrons and weaker repulsions between bonded pair and bonded pair electrons.
- Fri Nov 15, 2019 12:25 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Lone Pair Repulsion?
- Replies: 5
- Views: 378
Re: Lone Pair Repulsion?
Lone pair electrons are further from the positively-charged nucleus of an atom compared to bonding pairs of electrons and are therefore bound with less electrostatic force, causing them to have stronger repulsive forces than bonding pairs of electrons. Lone pairs have a larger range of negativity co...
- Fri Nov 15, 2019 12:00 pm
- Forum: Dipole Moments
- Topic: 3 Hydrocarbon Example?
- Replies: 1
- Views: 186
Re: 3 Hydrocarbon Example?
The 3 hydrocarbons at room temperature show that the greater amount of atoms there are, the stronger dispersion forces are, increasing the attractive forces between them, explaining why octadecane (C 18 H 38 ) is closer to a waxy solid than pentane (C 5 H 12 ), which is a mobile fluid, even though t...
- Fri Nov 15, 2019 11:51 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Lewis structure for 2E23 (a)
- Replies: 1
- Views: 100
Re: Lewis structure for 2E23 (a)
Yes, I believe there should be a double bond, but as VSEPR primarily looks at electron densities and the position of atoms, the lewis structure depicting the single bond between Sb and O allows one to predict the shape by primarily looking at bonded pairs of electrons and lone pairs of electrons.
- Fri Nov 15, 2019 11:41 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: induced dipoles
- Replies: 5
- Views: 351
Re: induced dipoles
Some examples of dipole-induced dipoles are HCl and N2, H2O and O2, and H2O and Br2
- Fri Nov 15, 2019 11:33 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Inter-molecular Electron Repulsion
- Replies: 6
- Views: 280
Re: Inter-molecular Electron Repulsion
The repulsion forces in order of least to most strong are:
bonding-bonding pr < lone-bonding pr < lone-lone pr
bonding-bonding pr < lone-bonding pr < lone-lone pr
- Fri Oct 04, 2019 12:36 pm
- Forum: Significant Figures
- Topic: Significant Figures
- Replies: 6
- Views: 416
Re: Significant Figures
It's best to apply significant figures at the end in order to retain as much accuracy as possible. Rounding early on or in the middle of calculations may initially seem convenient and more practical, but the dropping of a few decimal places/numbers can be severely detrimental to the final result, as...
- Fri Oct 04, 2019 12:26 pm
- Forum: Limiting Reactant Calculations
- Topic: General Limiting Reactant Question
- Replies: 4
- Views: 272
Re: General Limiting Reactant Question
There was only one reactant for M.3 and therefore, you did not need to find a limiting reactant/there was none. If there is more than one reactant, you would use the given moles/masses and molar masses to determine the limiting reactant and use that information to calculate the theoretical yield.
- Fri Oct 04, 2019 12:18 pm
- Forum: Limiting Reactant Calculations
- Topic: M5
- Replies: 2
- Views: 120
Re: M5
Just use the stoichiometric steps you used to calculate the the 12 mols of the ClO2F product to determine the moles of the other product (Br2).
- Fri Oct 04, 2019 11:56 am
- Forum: Empirical & Molecular Formulas
- Topic: E.11
- Replies: 1
- Views: 169
Re: E.11
Yes, you would need to consider the different masses of the isotopes because they have different percentages of abundance, both of which you would then use to calculate the mass of the average lithium atom by adding them together: (7.42/100)(9.988 * 10^-24 g) + (92.58/100)(1.165 * 10^-24 g) = the av...
- Fri Oct 04, 2019 11:38 am
- Forum: Molarity, Solutions, Dilutions
- Topic: Dilution Calculation
- Replies: 5
- Views: 238
Re: Dilution Calculation
Yes, you could technically leave the volume in mL since the molarity would cancel, but remember to keep in mind that otherwise, it would be wiser to leave it in liters.