Search found 78 matches
- Mon Mar 11, 2019 5:43 pm
- Forum: Ideal Gases
- Topic: Cell Diagram/Ecell
- Replies: 8
- Views: 946
Re: Cell Diagram/Ecell
The anode is the one that gets oxidized (loses electrons) and the cathode is the one that gets reduced (gains electrons)
- Mon Mar 11, 2019 5:25 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: 6.79 7th edition
- Replies: 1
- Views: 255
Re: 6.79 7th edition
I know it's mentioned as one of the hw problems but I wouldn't worry about it. I'm pretty sure he wouldn't put a problem that we wouldn't know how to do entirely.
- Mon Mar 11, 2019 5:17 pm
- Forum: Second Order Reactions
- Topic: 15.29 6th Edition
- Replies: 1
- Views: 248
15.29 6th Edition
15.29 For the first-order reaction A(s)-----> 3B + C, when [A] 0 0.015 mol/L, the concentration of B increases to 0.018 mol/L, in 3.0 min. (a) What is the rate constant for the reaction expressed as the rate of loss of A? (b) How much more time would be needed for the concentration of B to increase ...
- Thu Mar 07, 2019 7:07 pm
- Forum: Calculating Work of Expansion
- Topic: Keywords for each equation
- Replies: 4
- Views: 716
Re: Keywords for each equation
The unique reaction rate tells how how much the reactant is being used in that specific equation (reliant on the coefficients).
- Tue Mar 05, 2019 8:42 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Equation to rate graph
- Replies: 2
- Views: 374
Equation to rate graph
Hi! How would you use a balanced equation to create the rate of formation and use on a graph? For example, given 2A ----> B + 3C, how would you be able to visualize the graph that shows how fast A is used and C and B are made?
- Tue Mar 05, 2019 8:39 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: n in Nernst
- Replies: 4
- Views: 580
Re: n in Nernst
Hi! n is the moles of electrons transferred. Once you find the reduction and oxidation reactions, balance them. The coefficient in front of the electron value should be the same for the reduction and the oxidation reactions, and that is the n value.
- Tue Mar 05, 2019 8:38 pm
- Forum: Calculating Work of Expansion
- Topic: Keywords for each equation
- Replies: 4
- Views: 716
Re: Keywords for each equation
I know that I use the word "unique" to know if I have to use (-1/a)(dR/dt)
- Mon Feb 25, 2019 8:20 pm
- Forum: Balancing Redox Reactions
- Topic: Balancing Oxygens vs. Hydrogens
- Replies: 1
- Views: 338
Re: Balancing Oxygens vs. Hydrogens
No, but it makes the process easier. Chances are, you have a H+ that can balance out the excess H created by balancing the O, but you won't have an O2 to balance out the excess H.
- Mon Feb 25, 2019 8:18 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Vant Hoff Equation
- Replies: 3
- Views: 392
Re: Vant Hoff Equation
Hi! It isn't anything new in particular, it's just a combination of ∆G° = - RT ln K and ∆G° = ∆H° - T∆S°. You don't need to memorize it as long as you can make those equations equal to each other.
- Mon Feb 25, 2019 8:17 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Gibbs Units
- Replies: 2
- Views: 434
Re: Gibbs Units
Hi! Faraday's constant has C/mol and E has V which is J/C so you are left with J/m which is how G is written. You don't need to convert because the units work out anyways. The reason why the mole remains even after the moles of electrons are multiplied in is because if represented the energy release...
- Tue Feb 19, 2019 6:40 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: irreversible v reversible
- Replies: 4
- Views: 666
Re: irreversible v reversible
Hi! If it's a reversible process, you know that S has to change to something because the process cannot return to its original state.
- Tue Feb 19, 2019 6:34 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 9.63 Positive delta G
- Replies: 3
- Views: 650
Re: 9.63 Positive delta G
Hi! A negative dG of formation means that the G of reactants is greater than the G of products, so if free energy decreases, the end product is more stable. Lower energy means more stability.
- Mon Feb 18, 2019 7:37 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Positive vs Negative E value
- Replies: 5
- Views: 5610
Positive vs Negative E value
Hi! I had a quick question about determining how you know if the E cell is positive or negative. I know that it's positive if the cathode is right and negative if the cathode is on the left, but the electron flow always goes towards the cathode regardless of what side the cathode is on. I'm wonderin...
- Mon Feb 11, 2019 2:57 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Signs for entropy
- Replies: 5
- Views: 509
Re: Signs for entropy
Hi! You use the sign based on whatever your given ΔH is. If you aren't sure about that, you can use q and see if the temperature increased or decreased to see if the reaction is endothermic or exothermic. Endothermic reactions absorb heat, and the temperature increases so q is positive and ΔH is pos...
- Mon Feb 11, 2019 2:52 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 6th edition 9.61
- Replies: 1
- Views: 200
Re: 6th edition 9.61
Can you post the problem? Thanks!
- Mon Feb 11, 2019 2:49 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: Self-test 4I.4A
- Replies: 2
- Views: 567
Re: Self-test 4I.4A
So if you're going from gas to liquid, the reaction is H 2 O (g) -----> H 2 O (s) and if you wanted to write out the entropy of the reaction you could say it is the standard entropy of products - standard entropy of reactants. We're given those values: 196.9 J/molK - 86.8 J/molK = 110.1 J/molK and l...
- Tue Feb 05, 2019 9:33 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 4.D.11 (7th ed)
- Replies: 2
- Views: 333
Re: 4.D.11 (7th ed)
If you use PV = nRT you get 0.243 mol of N2. Using the standard enthalpy of formation of NO, you can say that the same amount of heat is released per mole of N2.
(0.243 mol)(180.6 KJ/mol) = 43.8 KJ
(0.243 mol)(180.6 KJ/mol) = 43.8 KJ
- Tue Feb 05, 2019 9:14 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: 4C.3 7th edition- calculate final temperature and change in enthaply
- Replies: 1
- Views: 909
Re: 4C.3 7th edition- calculate final temperature and change in enthaply
Hi! Yeah you're right you have to use qp = nCpdT but you also have to use Cp = (5/2) R. This comes from the equation and we didn't really cover it in class but it's what you use for a constant pressure situation. Calling the gas ideal allows us to use this specific heat capacity value.
- Tue Feb 05, 2019 9:09 pm
- Forum: Calculating Work of Expansion
- Topic: work done by system or on system
- Replies: 3
- Views: 489
Re: work done by system or on system
Hi! Here is a basic set of guidelines q is positive if system absorbs energy q is negative if system releases energy w is positive if work is being done on the system w is negative if work is being done by the system An example is going from solid to liquid; it requires heat to break the IMFs, so th...
- Mon Jan 28, 2019 1:14 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Enthalpy vs. Heat as a state property
- Replies: 1
- Views: 200
Re: Enthalpy vs. Heat as a state property
Hi! There some specifications for q=delH. The pressure has to be constant, so it's really qp=delH
- Mon Jan 28, 2019 1:11 pm
- Forum: Ideal Gases
- Topic: Problem 8.9 6th edition
- Replies: 1
- Views: 303
Problem 8.9 6th edition
Hi! I'm having trouble starting this problem off. Any help is appreciated! An ideal gas in a cylinder was placed in a heater and gained 5.50 kJ of energy as heat. If the cylinder increased in volume from 345 mL to 1846 mL against an atmospheric pressure of 750. Torr during this process, what is the ...
- Mon Jan 28, 2019 10:20 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: 8.3 6th Edition: Work being positive and negative
- Replies: 3
- Views: 329
8.3 6th Edition: Work being positive and negative
Hi! Can someone explain what work being positive or negative means, and how you can tell based off of the action that is being performed? This is the problem: 8.3 Air in a bicycle pump is compressed by pushing in the handle. If the inner diameter of the pump is 3.0 cm and the pump is depressed 20. c...
- Fri Jan 25, 2019 7:13 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 6th edition, 12.33
- Replies: 3
- Views: 327
Re: 6th edition, 12.33
Yes! The OH comes from the NaOH. If you wanted to write another equation, you could have Na2O+H2O --> OH + Na to show the inevitable dissociation of NaOH into Na and OH, but not Na2O+H2O --> NaOH + OH + Na because it kind of implies that NaOH does not completely dissociate.
- Wed Jan 23, 2019 10:31 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 6th edition, 12.33
- Replies: 3
- Views: 327
Re: 6th edition, 12.33
Hi! So using the 17.78 M OH from part b, you can use that and the balanced equation to work backwards. Because NaOH is a strong base, it will completely dissociate, meaning that [OH] = [Na] = [NaOH]. If we have 17.78 M [OH], the we have 17.78 M [NaOH]. The equation is Na2O + H2O ----> 2NaOH If we ha...
- Wed Jan 23, 2019 10:20 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Chemical Eq Part #3 Post-Module Assessment
- Replies: 1
- Views: 270
Re: Chemical Eq Part #3 Post-Module Assessment
So [C2H5OH] is 2 M and [CH3COOH] is 1 M and K is 4 so the ICE table looks like C2H5OH (aq) + CH3COOH (aq) ⇌ CH3COOC2H5 (aq) + H2O (l) 2 M 1 M 0 M -x -x +x 2-x 1-x x The equation is (x)/((2-x)(1-x)) = 4 and because K isn't that small, you have to do the whole quadratic and all that. x/(2-3x+x^2) = 4 ...
- Mon Jan 21, 2019 3:36 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 6th Edition Hw #12.69
- Replies: 1
- Views: 234
Re: 6th Edition Hw #12.69
Hi! I think this goes back to the coordination compounds that we learned about in 14A. When Al forms coordinate covalent bond, it bonds with 6 other waters.
- Mon Jan 14, 2019 9:19 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Autoprotolysis
- Replies: 3
- Views: 322
Re: Autoprotolysis
I think it works for anything as long as it follows that guideline
- Mon Jan 14, 2019 9:18 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: 12.23 in sixth edition
- Replies: 1
- Views: 212
Re: 12.23 in sixth edition
Given Kw of 2.1 10^-14, you know that [H3O+] and [OH-] have to be the same because water is neutral so one cannot be greater than the other. So the [H3O+] and [OH-] is the square root of Kw
- Mon Jan 14, 2019 4:33 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: 11.27
- Replies: 2
- Views: 107
Re: 11.27
Hi! Set up the reaction to solve for Q. Given the equation PCl5(g) <----> PCl3(g) + Cl2(g), Q = ((PPCl3)(PCl2))/(PPCl5)). You're given PPCl5 = 1.18 bar, PCl2 5.43 bar, and K=25. You set Q=K and plug in what you have.
[(PPCl3)(5.43)]/(1.18) = 25
PPCl3 is 5.43 bar
[(PPCl3)(5.43)]/(1.18) = 25
PPCl3 is 5.43 bar
- Tue Jan 08, 2019 10:46 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Leaning to the left/right
- Replies: 4
- Views: 791
Re: Leaning to the left/right
Left means reactants and right means products. For example in the formation of ammonia 2NH3 <----> N2 + 3H2 If the reaction is left leaning, then it tends to want to make more ammonia. If the reaction is right leaning, it tends to make more nitrogen gas and hydrogen gas. Increasing the concentration...
- Tue Jan 08, 2019 10:41 pm
- Forum: Student Social/Study Group
- Topic: New to Lavelle
- Replies: 32
- Views: 5345
Re: New to Lavelle
Honestly I just do homework problems and the audio visual practice (you find it on his website). His UAs are undergraduate assistants that host small review sessions. There are: Step ups for people that struggle a bit with chem, it's for them to start from the basics and work their way up Drop ins f...
- Tue Jan 08, 2019 10:36 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Discussion Points
- Replies: 3
- Views: 244
Re: Discussion Points
Hi! As long as you post 3 times a week in any fashion (answering questions, posting, responding in discussion), you get points! The way to check is to revisit a thread that you have posted on and underneath your name, you'll see how many times you've posted. You can also check in the section called ...
- Mon Dec 03, 2018 8:49 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: CO3 2-
- Replies: 2
- Views: 388
Re: CO3 2-
Hi! It can't be tridentate because CO3 2- is trigonal planar, so the molecule can't reorient itself to bond all three oxygens to the central atom. However, one or two oxygens can bind to the central atom because the molecule won't need to readjust to let that happen.
Re: Order/OH2
Hi! They write OH2 to indicate that it's the O that bonds. So I think you have to write it like OH2. I think you write them in alphabetical order?? But I could be wrong about that.
- Mon Dec 03, 2018 8:44 pm
- Forum: Bronsted Acids & Bases
- Topic: No Proton Transfer?
- Replies: 2
- Views: 317
Re: No Proton Transfer?
Hi! Ideally, if NH3 was a Bronsted base, it would accept the H+ from CH3COOH. But because CH3COOH does not donate a proton, it is not a Bronsted acid, so NH3 cannot be a Bronsted base. There isn't a clear cut transfer of protons. If this were a Bronsted problem, the reaction would look like: CH3COOH...
- Mon Nov 26, 2018 7:31 pm
- Forum: Sigma & Pi Bonds
- Topic: Resonance structure
- Replies: 2
- Views: 246
Re: Resonance structure
Hi! It should have the same number of sigma and pi bonds regardless of whether or not the molecule has resonance. Otherwise the structures would not be equal. Because SO3-2 has two single bonds and one double bond, here are three sigma bonds (one per single bond and one in the double bond) and one p...
- Mon Nov 26, 2018 2:41 pm
- Forum: Hybridization
- Topic: Bond Angles and Hybridization
- Replies: 2
- Views: 368
Re: Bond Angles and Hybridization
So for sp, the bond angle is 180. We can say (sort of) that 50% of this is contributed from the s and 50% is contributed from the p (because sp is one half s and one half p). For sp2, the bond angle is 120. Now about 33% of the angle is from the s and 67% is from the p (sp2 is one third s and two th...
- Mon Nov 26, 2018 1:44 pm
- Forum: Polarisability of Anions, The Polarizing Power of Cations
- Topic: 2D.3
- Replies: 2
- Views: 263
Re: 2D.3
You only need to look at the trends. I don't think we get access to actual values, so we use trends to estimate which compounds have higher ionic character.
- Tue Nov 20, 2018 8:21 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: 2E.25 (7th Ed.) Lewis structure and polarity
- Replies: 2
- Views: 288
Re: 2E.25 (7th Ed.) Lewis structure and polarity
So you're right, but when you think about its molecular geometry, it makes sense why it's at least a little polar. CH2Cl2 is tetrahedral, and when you arrange the atoms around C in the tetrahedral shape, you can never orient them so that the charges fully cancel. Nothing is directly across from each...
- Tue Nov 20, 2018 8:14 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Help w/ 4.19 (6th edition)
- Replies: 2
- Views: 311
Re: Help w/ 4.19 (6th edition)
Be has 4 electrons total, and from that has 2 valence electrons. If of course cannot follow the octet rule because it doesn't have enough electrons. So, it can only form one bond with each carbon which uses up its two valence electrons. The structure will look like 3HC-Be-CH3 but the Hs will be spre...
- Tue Nov 20, 2018 8:10 am
- Forum: Polarisability of Anions, The Polarizing Power of Cations
- Topic: Polarizability and Ionic Character
- Replies: 2
- Views: 1183
Re: Polarizability and Ionic Character
If an ion has more polarizing power, that means that it is able to influence another atom's electron cloud shape. For example, Ca2+ has polarizing power, and can alter the electron cloud of Cl- (or other ions). The smaller the size of the ion, and the larger the charge, the more polarizing power a c...
- Tue Nov 20, 2018 8:06 am
- Forum: Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)
- Topic: hydrogen bonding
- Replies: 11
- Views: 1008
Re: hydrogen bonding
Hi! Hydrogen bonding is stronger than most intermolecular forces, so it is requires more energy to break those IMFs, and therefore needs a higher temperature for the compound to move from solid to liquid. I thought covalent bonds and ionic bonds were stronger? Or do those not count as IMF? Covalent...
- Wed Nov 14, 2018 2:53 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: VSEPR Formula
- Replies: 1
- Views: 248
Re: VSEPR Formula
Hi! I think they're asking for the AXE format, where A is the central atom, X are the atoms that bond to the central, and E is the number of lone pairs on the central atom. For example, SCl4 has one central atom, four surrounding atoms, and one lone pair on the central atom So the VSEPR formula is A...
- Wed Nov 14, 2018 9:06 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Bond Angles
- Replies: 4
- Views: 434
Re: Bond Angles
Yeah, I think we need to memorize bond angles for the major molecular geometry shapes (180, 120, 109.5, 90/120, 90) as you move from linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral, and then just put ranges for anything with lone pairs. Eg. Trigonal planar: 120, but Bent could...
- Tue Nov 13, 2018 11:15 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: Lewis Structures before Forces?
- Replies: 2
- Views: 299
Re: Lewis Structures before Forces?
I think until you can intuitively figure it out, it's best to draw the Lewis structures. For me, unless I'm 100% sure of its structure (like CH4), I will draw the full Lewis structure.
- Tue Nov 13, 2018 2:54 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: hydrogen bonding
- Replies: 11
- Views: 1008
Re: hydrogen bonding
Hi! Hydrogen bonding is stronger than most intermolecular forces, so it is requires more energy to break those IMFs, and therefore needs a higher temperature for the compound to move from solid to liquid.
- Fri Nov 09, 2018 1:08 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: Liquid and Solid Formation
- Replies: 2
- Views: 376
Re: Liquid and Solid Formation
The stronger the dispersion forces, the stronger the force between the molecules. The stronger the intermolecular forces (IMFs) are, the more solid the compound becomes (except for H20, which has stronger IMF as a liquid vs. solid, which explains why water is denser than ice). IMFs will form between...
- Mon Nov 05, 2018 1:32 pm
- Forum: Lewis Structures
- Topic: Lewis Structures for Ionic Bonds
- Replies: 5
- Views: 524
Re: Lewis Structures for Ionic Bonds
There's not really a difference when drawing, because Lewis structures just outline the lone pairs and what types of bonds exist within a molecule.
- Mon Nov 05, 2018 1:24 pm
- Forum: DeBroglie Equation
- Topic: Homework 1B 21
- Replies: 1
- Views: 447
Re: Homework 1B 21
Hi! I have 6th edition but I think you're talking about the baseball problem. You use De Broglie's equation for this problem. m would equal the mass, and you need to convert 5.15 oz to kilograms, which is 145.745 x 10^3 g. If you convert 92 mph to m/s you get 41.128 m/s as the velocity. You are give...
- Mon Nov 05, 2018 1:10 pm
- Forum: Resonance Structures
- Topic: Radicals
- Replies: 11
- Views: 1429
Re: Radicals
Hi! A radical is when an atom has one unbonded electron that is not in a pair. For example, in CH3, C has a radical because its fourth electron is not bonded. You draw your Lewis structure normally, and just put one dot instead of two to indicate a radical.
- Sun Nov 04, 2018 12:33 pm
- Forum: Resonance Structures
- Topic: 2B.15 (7th ed) [ENDORSED]
- Replies: 1
- Views: 177
Re: 2B.15 (7th ed) [ENDORSED]
Hi! You are allowed to have one double bond (between the N and O), but you cannot have two for both N O bonds. Because N is not in the 3rd period, it cannot make an expanded octet using the empty 3d orbital. If it could, then it could have five bonds (two double bonds to O and one single bond to Cl)...
- Thu Nov 01, 2018 10:06 pm
- Forum: Electronegativity
- Topic: What are the trends useful for?
- Replies: 12
- Views: 1018
Re: What are the trends useful for?
You could use IE to determine if a bond has more ionic character or covalent character. The greater the difference between IE between the elements, the more ionic character is has. The others have uses too, but I can't think of any right now.
- Tue Oct 30, 2018 3:26 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: configuration for ions
- Replies: 2
- Views: 322
Re: configuration for ions
So F normally is [He] 2s2 2p5 but because you have F-, you add another electron. This electron would go to the p orbital (becuase 2p5 still has space for one more electron), and then you have [He] 2s2 2p6 which is the same as [Ar].
- Tue Oct 30, 2018 3:22 pm
- Forum: Octet Exceptions
- Topic: central atoms with more than 4 bonds
- Replies: 3
- Views: 507
Re: central atoms with more than 4 bonds
S has 6 valence electrons so it's allowed to make two double bonds (takes 4 electrons) and two single bonds (takes 2 electrons). You want the central atom to have as few lone pairs as possible for stability purposes, that's why S makes some double bonds too
- Mon Oct 29, 2018 10:50 pm
- Forum: Lewis Structures
- Topic: lewis structures
- Replies: 3
- Views: 211
Re: lewis structures
Some of the exception are H, He, Li, Be because they don't have enough electrons to even make the octet. Sulfur, chlorine, silicon and phosphorus can all go beyond the octet, so they are also exceptions. There are some more for sure, but Br follows the octet rule (someone please correct me if I'm wr...
- Mon Oct 29, 2018 3:11 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Do non-science questions count as posts?
- Replies: 2
- Views: 241
Re: Do non-science questions count as posts?
Hi! I'm sure if you talk to your TA, they can let you know if you got credit for it.
- Mon Oct 22, 2018 8:10 pm
- Forum: Properties of Electrons
- Topic: Atomic Spectra
- Replies: 1
- Views: 320
Re: Atomic Spectra
Okay I think you would start off by dividing 1 meter by 1,650,763.73 wavelengths to get the length of one wavelength of Krypton-86. This makes sense because the units become meters/one wavelength and that's what wavelength measures. 1/1,650,763.73 = 6.058 x 10-7 m which is 605.8 nm. That is in the v...
- Mon Oct 22, 2018 6:08 pm
- Forum: Accuracy, Precision, Mole, Other Definitions
- Topic: Group 5 Transition Metals
- Replies: 1
- Views: 585
Re: Group 5 Transition Metals
Hi! Dr. Lavelle posted #55c as an error in the solution's manual for 6th edition, so I'm assuming the same happened for 7th. He says that it should be (n-1)d3 ns2. So you're right, the group 5 transition metals start with Vanadium and so on.
- Mon Oct 22, 2018 6:04 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Electron configuration for elements > Z = 56
- Replies: 3
- Views: 736
Re: Electron configuration for elements > Z = 56
Yeah I think you're right I was going through lecture notes and he mentioned that today. Thank you!
- Mon Oct 22, 2018 4:01 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Electron configuration for elements > Z = 56
- Replies: 3
- Views: 736
Electron configuration for elements > Z = 56
Hi! Do we need to include the f orbital when finding the electron configuration of Tungsten (z=74) for example? Would it be
[Xe] 4f14 5d4 6s2 or [Xe] 6s2 5d4
Also if it is the first one, in what order should the orbitals be listed? Thank you!
[Xe] 4f14 5d4 6s2 or [Xe] 6s2 5d4
Also if it is the first one, in what order should the orbitals be listed? Thank you!
- Wed Oct 17, 2018 7:34 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Orbitals
- Replies: 1
- Views: 92
Re: Orbitals
I totally agree, I think it should be 4s2, 3d10 just because the periodic table moves in that progression. I think as long as you include everything, it shouldn't matter!
- Tue Oct 16, 2018 10:42 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Formulas?
- Replies: 4
- Views: 435
Re: Formulas?
He also mentioned that he really does not want us using Rhydberg's equation because there's more room for error. He wants us to use E = -hR/(n^2) for each energy level and then go E(final)-E(initial) so that we actually understand the problem and don't get positive and negative signs mixed up. Someo...
- Tue Oct 16, 2018 10:10 pm
- Forum: DeBroglie Equation
- Topic: Homework Problem 1b.15
- Replies: 2
- Views: 325
Re: Homework Problem 1b.15
Hi! I think you mean mass of electron. Photons have no mass, so because this question asks "what is the wavelength of the ejected electron" you use De Broglie's equation. The mass of an electron is always the same, regardless of the element. I think it's 9.11 x 10^-31 kg and you have to le...
- Tue Oct 16, 2018 10:07 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: when to use which formula
- Replies: 1
- Views: 156
Re: when to use which formula
I think you use the first equation when finding the energy change from one energy level to another, and the second equation when the question specifically mentions a side length, or something like "box".
- Sun Oct 14, 2018 5:19 pm
- Forum: Properties of Light
- Topic: Visible Light Spectrum
- Replies: 3
- Views: 238
Re: Visible Light Spectrum
Hi! I know that we should know that 400 nm is about violet and 700 nm is about red, but I don't think we need to be able to match color to wavelength exactly.
- Fri Oct 12, 2018 8:44 am
- Forum: Properties of Light
- Topic: Homework Problem #7
- Replies: 1
- Views: 263
Re: Homework Problem #7
(a) Using the equation c = λ * v (λ is wavelength, v is frequency, c is the speed of light), you can sub in what you know! 3.0 x 10^8 m/s = (7.1 x 10^14 s^-1) x (λ) λ = 4.225 x 10^-7 m and they want it written in nanometers so it become 422.5 nm. And because 422.5 nm falls on the violet light wavele...
- Fri Oct 12, 2018 8:31 am
- Forum: Empirical & Molecular Formulas
- Topic: F13
- Replies: 3
- Views: 730
Re: F13
I think it's just like basic compounds, like knowing that CO3 is carbonate so Na2CO3 is sodium carbonate. If you looked up "polyatomic ions", those are probably the only ones that you would need to know. There's only like 6 I think that you would really need to know. Carbonate CO3 (-2) Nit...
- Tue Oct 09, 2018 8:34 am
- Forum: Molarity, Solutions, Dilutions
- Topic: Edition 6, G5 Question
- Replies: 1
- Views: 227
Re: Edition 6, G5 Question
I think that's a typo!
- Mon Oct 08, 2018 9:29 pm
- Forum: Limiting Reactant Calculations
- Topic: Question M17 from 6th Edition
- Replies: 2
- Views: 414
Re: Question M17 from 6th Edition
Thank you!
- Mon Oct 08, 2018 8:58 pm
- Forum: Limiting Reactant Calculations
- Topic: Question on Post-Module #22
- Replies: 2
- Views: 228
Re: Question on Post-Module #22
Hi! I think the equation wasn't fully balanced. This is what is given: C6H9Cl3 + 3AgNO3 ---> AgCl + C6H9(NO3)3. But there are 3 moles of Ag on the left and only 1 mole on the right. There should be a 3 in front of the AgCl. That means if you multiply your 0.5732 mol by 3, you get 1.72 g and that is ...
- Mon Oct 08, 2018 8:46 pm
- Forum: Molarity, Solutions, Dilutions
- Topic: Finding the volume of stock solution to dilute
- Replies: 13
- Views: 1360
Re: Finding the volume of stock solution to dilute
Hi! I don't think it really matters, but I tend to convert to liters if the value exceeds the thousands place. So I would convert 3,000 ml to 3 L just to keep it shorter. I think a TA said that it doesn't matter for the test!
- Mon Oct 08, 2018 8:43 pm
- Forum: Limiting Reactant Calculations
- Topic: Question M17 from 6th Edition
- Replies: 2
- Views: 414
Question M17 from 6th Edition
Hi! I was having some trouble with this problem. I have found the molar mass of XA to be 338 g/mol, but I'm not sure where to go from there. The acid HA (where A stands for an unknown group of atoms) has molar mass 231 g/mol. HA reacts with the base XOH (molar mass 125 g/mol) to produce H2O and the ...
- Sun Oct 07, 2018 9:15 am
- Forum: Limiting Reactant Calculations
- Topic: Limiting Reactants M9 [ENDORSED]
- Replies: 2
- Views: 372
Re: Limiting Reactants M9 [ENDORSED]
Nitrate is NO3, and has a charge of -1 Copper is Cu, and is given as +2 (as indicated by the (II)) Sodium is Na, and has a charge of +1 Hydroxide is 0H, and has a charge of -1 Copper (II) nitrate would be Cu(NO3)2 and sodium hydroxide would be NaOH. The products would be NaNO3 and Cu(OH)2. The react...
- Fri Oct 05, 2018 11:03 pm
- Forum: Balancing Chemical Reactions
- Topic: H19
- Replies: 3
- Views: 384
Re: H19
You're right, but I think because the aspartame has nitrogen in it, the gas needs to escape as a product somehow. I think that in general, all combustion reactions have CO2 and H2O as products, but might sometimes include other products to make up for any elements in the reactant that is not hydroge...
- Thu Oct 04, 2018 6:28 pm
- Forum: Limiting Reactant Calculations
- Topic: Mass of products/Reactants [ENDORSED]
- Replies: 3
- Views: 252
Re: Mass of products/Reactants [ENDORSED]
The mass of the products should equal the mass of the reactants. There can be more moles of product than reactant or vice versa, but there cannot be different masses.
- Tue Oct 02, 2018 8:24 am
- Forum: Empirical & Molecular Formulas
- Topic: Question F11 from 6th edition
- Replies: 3
- Views: 1806
Re: Question F11 from 6th edition
Thank you! I think I rounded a little bit too much so my values weren't as close to a whole number.
- Tue Oct 02, 2018 8:03 am
- Forum: Molarity, Solutions, Dilutions
- Topic: How to find volume of solution w specific amount of Na2CO3?
- Replies: 2
- Views: 3193
Re: How to find volume of solution w specific amount of Na2CO3?
From part (a) and (b) we know that the original concentration of the sodium carbonate (2.111 g) is .07966 M. This comes from converting grams of sodium carbonate to moles, and then dividing that by the total volume in liters (2.111 g) / (106 g) = 0.0199 mol Na2CO3 / 0.250 Liters = 0.07966 M Na2CO3 T...
- Tue Oct 02, 2018 7:49 am
- Forum: Empirical & Molecular Formulas
- Topic: Question F11 from 6th edition
- Replies: 3
- Views: 1806
Question F11 from 6th edition
Hello! I am struggling to finish the last steps of this problem: "The mass percentage composition of cryolite, a compound used in the production of aluminum, is 32.79% Na, 13.02% Al, and 54.19% F." I am asked to find the empirical formula. I am fine up until after I have divided the three ...