Search found 69 matches
- Wed Mar 13, 2019 5:11 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: exercise 11.41
- Replies: 1
- Views: 223
Re: exercise 11.41
Concentration is calculated by the equation moles/liters, so as long as you find out how many moles there are of each gas and divide by the volume of the container, in this case a 0.250 L flask, then you will be able to find the concentration of the gases. You are given the mass of CO2 gas at equili...
- Wed Mar 13, 2019 4:44 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: [tex]\Delta U = \Delta H[/tex]
- Replies: 2
- Views: 924
Re: [tex]\Delta U = \Delta H[/tex]
The two equations you have there are essentially the same equations derived from delta(U) = delta(H) + w where w is work. Since w = -Pdelta(V) where P is pressure and delta V is the change in volume, we can rewrite the equation as delta(U) = delta(H) - Pdelta(V). By the ideal gas law, PV = nRT, -Pde...
- Wed Mar 13, 2019 4:35 pm
- Forum: Second Order Reactions
- Topic: 15.37
- Replies: 1
- Views: 301
Re: 15.37
This problem states that it is a first order decomposition of SO2Cl2, so you know that the only reactant is SO2Cl2 because that's what decomposition means. All parts of the question don't actually require the use of a complete chemical equation for the decomposition, so you don't actually need to fi...
- Wed Mar 06, 2019 5:22 pm
- Forum: General Rate Laws
- Topic: Mechanisms?
- Replies: 3
- Views: 405
Re: Mechanisms?
Not sure yet, but Dr. Lavelle said he'd be talking about it in lecture on Friday
- Wed Mar 06, 2019 5:05 pm
- Forum: First Order Reactions
- Topic: How to do 15.11 6th edition?
- Replies: 4
- Views: 549
Re: How to do 15.11 6th edition?
The problem tells you that N2O5 decomposes by a first order reaction, so the rate law will have the form rate = k[A] where A is the reactant, N2O5, and the it is raised to the first power since it is a first order reaction. You are also given k = 5.2 x 10^-3. As mentioned before, you don't need to d...
- Wed Mar 06, 2019 3:34 pm
- Forum: Zero Order Reactions
- Topic: Reaction orders
- Replies: 2
- Views: 454
Re: Reaction orders
Dr. Lavelle said in class that we won't have to worry about third order reactions, but for first and second order reactions, you can plot the data given to you on a graph or examine the rate law. If a graph of ln[A] vs. t shows a straight line with a negative slope, you know that it is a first order...
- Wed Feb 27, 2019 2:46 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.11d 6th edition
- Replies: 1
- Views: 225
Re: 14.11d 6th edition
the half reaction for the anode here should be H2O --> O2 + H+ and you first balance the O's and then H's and then the charge to get 2H2O --> O2 + 4H+ + 4e- ; this way you see that the O has been oxidized
- Wed Feb 27, 2019 2:41 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Pt (s) [ENDORSED]
- Replies: 8
- Views: 970
Re: Pt (s) [ENDORSED]
you do not need to add Pt(s) to liquids in cases where you have Hg(l) and Hg ions, but I am not sure if there are any exceptions/other cases
- Wed Feb 27, 2019 2:34 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.11 6th edition, part D
- Replies: 2
- Views: 327
Re: 14.11 6th edition, part D
I think the solution manual is correct in saying that O2 and H+ are the reactants in the oxidation half reaction because there is no way to get H+ ions from O2 gas, so they must react together to form water
- Wed Feb 20, 2019 3:42 pm
- Forum: Balancing Redox Reactions
- Topic: Separating the equation
- Replies: 5
- Views: 510
Re: Separating the equation
by separating the equation, Dr. Lavelle means that you should separate the oxidation part from the reduction part for example, if you have the equation Cu2+ + Zn --> Cu + Zn2+ you can separate the Cu's and Zn's to get Cu2+ + 2e- --> Cu (reduction) and Zn --> Zn2+ + 2e- (oxidation) (you don't see the...
- Wed Feb 20, 2019 3:38 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Cathode vs Anode
- Replies: 1
- Views: 222
Re: Cathode vs Anode
the cathode will contain the reduction reaction and the anode will contain the oxidation reaction. something I find helpful to help me remember is "red cat" for reduction in cathode and "an ox" for anode = oxidation
- Wed Feb 20, 2019 3:37 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Entropy and mass
- Replies: 2
- Views: 540
Re: Entropy and mass
usually, a more massive atom will have greater entropy because that means it has more protons, neutrons, and electrons than anther atom with less mass, and because it has more particles, there are more possible states it could be in, increasing its total entropy
- Thu Feb 14, 2019 2:36 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Cp,m
- Replies: 3
- Views: 717
Re: Cp,m
Cp,m is for an ideal monoatomic gas at constant pressure and Cv,m is for an ideal monoatomic gas at constant volume. You use these values usually when you are trying to calculate the change in entropy when there is a temperature change.
- Thu Feb 14, 2019 2:34 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Reversible vs. Irreversible
- Replies: 5
- Views: 1051
Re: Reversible vs. Irreversible
For a reversible reaction, the system is at equilibrium and when it is either compressed or expanded, it does so slowly in infinitely small increments, and you use the equation w = -nRTln(V2/V1) for work. For an irreversible reaction, the reaction is happening under a constant external pressure Pext...
- Thu Feb 14, 2019 2:30 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Expansion of a system
- Replies: 8
- Views: 1198
Re: Expansion of a system
If a system is doing work, it is usually expanding unless the problem specifically states that the system is at constant volume.
- Mon Feb 04, 2019 11:47 pm
- Forum: Calculating Work of Expansion
- Topic: 6th ed 8.27
- Replies: 1
- Views: 217
Re: 6th ed 8.27
for part a), when you plugged in the numbers into w = -PdV you get -2.23 atm x L (single digits) but work is in terms of joules, so you need to convert that into joules using 1 atmL = 101.325 J, which gives you -226 J for your answer the same thing needs to be done for part b) unless for R you use 8...
- Mon Feb 04, 2019 11:41 pm
- Forum: Phase Changes & Related Calculations
- Topic: HW 8.41 6th edition
- Replies: 2
- Views: 361
HW 8.41 6th edition
when calculating the heat of the ice cube, why do we use the specific heat of water in its liquid form and not its solid form?
- Mon Feb 04, 2019 11:39 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Purpose of ln
- Replies: 2
- Views: 257
Re: Purpose of ln
I'm assuming you're referring to this equation w = - ∫ PdV (limits of integration from V1 to V2) = -nRT ln (V2/V1), in which case the ln comes from taking the integral of 1/V when you replace P with nRT/V (ideal gas equation rearranged) and the constants are taken out so you're left with ∫ dV/V. Sin...
- Fri Feb 01, 2019 12:00 am
- Forum: General Science Questions
- Topic: Redox reaction
- Replies: 1
- Views: 661
Re: Redox reaction
Gaining an H atom does not mean gaining an H+ ion since an H atom still has an e-; this is an important distinction to make because when that new H atom is gained, its one e- will spend more time around the central atom it bonded to because it is more electronegative, and thus, that central atom is ...
- Thu Jan 31, 2019 11:49 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.65 6th Edition
- Replies: 2
- Views: 310
Re: 8.65 6th Edition
since the standard enthalpy of formation is the delta H for the formation of one mole of a substance from its elements in their most stable form, by definition, the standard enthalpy of formation of elements in their most stable form is = 0. the most stable form of oxygen is O2, so since the delta H...
- Thu Jan 31, 2019 11:45 pm
- Forum: Phase Changes & Related Calculations
- Topic: 8.1 (6th Edition)
- Replies: 1
- Views: 259
Re: 8.1 (6th Edition)
open systems are usually exposed to the air and can exchange both matter and energy: in this case, the gas burning in a car engine (releases heat, which is part of energy and exhaust which contains matter) and a living plant (exchanges gases like O2 and CO2 which counts as matter) are open systems c...
- Thu Jan 24, 2019 12:34 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Chemical Equilibrium Post Assessment part 3 q20
- Replies: 2
- Views: 289
Re: Chemical Equilibrium Post Assessment part 3 q20
Since Kc = 5.66 x 10^-10 <<< 10^-3, you can disregard the change in molarity caused by x for SO2 and O2, so instead of having Kc = (2x)^2/(0.522-x)^2(0.633-x), you'll just have Kc = (2x)^2/(0.522)^2(0.633), which you won't even have to use the quadratic equation for.
Answer should be D.
Answer should be D.
- Wed Jan 23, 2019 6:53 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Sig Figs when dealing with the change in concentration value
- Replies: 2
- Views: 259
Re: Sig Figs when dealing with the change in concentration value
it's always best to save rounding for the end, and this could mean either the equilibrium constant value or the x-value, depending on what the question is asking for; basically, only round for whichever number is your final answer.
- Wed Jan 23, 2019 5:52 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 6th Ed. 11.57
- Replies: 1
- Views: 200
Re: 6th Ed. 11.57
In this problem, the water in the equation is in its gaseous state, so we need to include it in the ICE table. The only time we don't include water is when it is a liquid/is acting as a solvent.
- Thu Jan 17, 2019 11:46 am
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: pKa and pKb
- Replies: 4
- Views: 950
Re: pKa and pKb
the same trend also applies for pKb: the lower the pKb, the stronger the base
- Thu Jan 17, 2019 11:44 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Changes to K value
- Replies: 9
- Views: 720
Re: Changes to K value
You could also potentially change K by multiplying the reaction equation by a whole number. If you double the stoichiometric coefficients of everything in the equation, the K will become K^2, which will either increase K is K > 1 or decrease K if K < 1.
- Thu Jan 17, 2019 11:41 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: problem 11.11 6th edition
- Replies: 1
- Views: 119
Re: problem 11.11 6th edition
You are given that ozone forms oxygen in the equilibrium reaction 2O3 -->/<-- 3O2. It gives you different conditions, one where 0.10 mol of ozone is put in a 1.0L vessel and then another where 0.50 mol is put in a 1.0L vessel, both of which are at the same temperature. This means that the equilibriu...
- Thu Jan 10, 2019 3:19 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: K, Kc and Kp
- Replies: 8
- Views: 604
Re: K, Kc and Kp
I think the way the textbook explains it is that K and Kp are essentially the same thing, using partial pressures for reactions that have gaseous compounds while Kc is specific to a K that uses concentrations.
Pages 424 and 437 in the 6th edition might be helpful if you want to look at it.
Pages 424 and 437 in the 6th edition might be helpful if you want to look at it.
- Thu Jan 10, 2019 2:56 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 6th edition, 11.7?
- Replies: 4
- Views: 346
Re: 6th edition, 11.7?
What's going on in the flask is the reaction X2 --> 2X, and normally, you'd just use the equation K = (Px)^2/(Px2). However, for this, I used the partial pressures of each gas since it only gives you the pressure of X2 at the very beginning, so you modify the equation to be K = (nx*Pi)^2/(nx2*Pi) wh...
- Wed Jan 09, 2019 5:51 pm
- Forum: Ideal Gases
- Topic: post assessment questions
- Replies: 2
- Views: 371
Re: post assessment questions
since the given chemical equation is all gases, you can rearrange the pv = nrt equation to p = nrt/v = Mrt (where M is molarity for concentration since n/V is mol/L) to find the partial pressures of the gases from the given concentrations; then, you can find the equilibrium constant using P(products...
- Wed Dec 05, 2018 2:42 pm
- Forum: Conjugate Acids & Bases
- Topic: Conjugate acid base example from class
- Replies: 2
- Views: 877
Re: Conjugate acid base example from class
CH3CHO2- is the base and CH3CO2H is its conjugate acid (first pair). H2O is the acid and OH- is its conjugate base (second pair).
- Tue Dec 04, 2018 11:32 pm
- Forum: Bronsted Acids & Bases
- Topic: Water as an Acid or Base
- Replies: 4
- Views: 543
Re: Water as an Acid or Base
Whether H2O will act as an acid or base depends on the nature of the compound it is reacting with. For example, if water were to react with ammonia (a weak base), then it would be an acid: NH3 + H2O --> NH4+ + OH- (water is donating its proton and ammonia is accepting it). If water were to react wit...
- Tue Dec 04, 2018 2:36 pm
- Forum: Naming
- Topic: 7th Edition 9C. 1
- Replies: 3
- Views: 321
Re: 7th Edition 9C. 1
They're essentially the same thing. Cyano is just the older naming convention and cyanido is the newer name agreed on by the IUPAC. I think you can use either one and be correct since the book said the old names are still heavily in use.
- Thu Nov 29, 2018 10:21 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Charges
- Replies: 4
- Views: 526
Re: Charges
You know that molecules like ammonia (NH3) have a zero charge and compounds like cyanide (CN-) have charges on them. Add the charge of the compound/molecule to the oxidation state of the anion (i.e. -1 for Cl). Then, you look at the overall charge of the coordinate compound to figure out the oxidati...
- Thu Nov 29, 2018 3:04 pm
- Forum: Hybridization
- Topic: Rating Polarity
- Replies: 3
- Views: 326
Re: Rating Polarity
If all those atoms in the same group where bonded to the same atom, let's say the halogen group are all bonded to hydrogen atoms (HF, HCL, HBR, HI), the atom that will form a more polar bond would be the most electronegative one, which is the first atom in each group. This is because it has the leas...
- Thu Nov 29, 2018 2:59 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: radical
- Replies: 5
- Views: 663
Re: radical
A radical is an atom or a molecule that has an unpaired valence electron such as fluorine or CH3. Radicals are generally very unstable and can cause harm to the body by denaturing macromolecules.
- Wed Nov 21, 2018 12:29 am
- Forum: Sigma & Pi Bonds
- Topic: formation of pi bond
- Replies: 6
- Views: 639
formation of pi bond
Can a pi bond be formed before a sigma bond is formed?
- Wed Nov 21, 2018 12:27 am
- Forum: Hybridization
- Topic: Forming the different types of bonds
- Replies: 2
- Views: 301
Forming the different types of bonds
Why does a p-p orbital overlap form a sigma bond and not a pi bond? What is an example of this?
- Wed Nov 21, 2018 12:22 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: predicting dipole moments
- Replies: 4
- Views: 428
Re: predicting dipole moments
I would first begin by looking at the nature of the bonds in the molecule and then the molecule shape because the intermolecular forces formed depend on the polarity of the molecule. If the molecule consists entirely of nonpolar bonds, then the molecule itself will be nonpolar as well, meaning it wi...
- Wed Nov 14, 2018 3:14 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Homework Question 4.1 (6th edition)
- Replies: 4
- Views: 836
Re: Homework Question 4.1 (6th edition)
You're right! The shape wouldn't be linear if there was a/were lone pair(s) on the central atom. That's why for a) where the bond angle is 120 degrees, it is possible for there to be lone pairs on the central atom since the shape of the molecule is bent (repulsion from the lone pair(s) push the two ...
- Tue Nov 13, 2018 10:44 pm
- Forum: Polarisability of Anions, The Polarizing Power of Cations
- Topic: Determining the polarizability
- Replies: 4
- Views: 563
Re: Determining the polarizability
You don't really look at the charge of the anion when determining its polarizability but rather its size and electronegativity. Polarizability increases as the ion gets larger and less electronegative since the electrons aren't held as tightly by the nucleus. One of the homework problems shows this ...
- Tue Nov 13, 2018 9:38 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: 2
- Views: 279
hydrogen bonding
During lecture, Dr. Lavelle said that the partial positive charge on a hydrogen interacts with the/a lone pair on an N, O, or F. I always thought the interaction was with an N, O, or F that had a partial negative charge. Is this incorrect or do both occur at the same time when a hydrogen bond forms?...
- Tue Nov 06, 2018 8:41 pm
- Forum: Resonance Structures
- Topic: bond lengths of resonant atoms
- Replies: 4
- Views: 572
Re: bond lengths of resonant atoms
here's an example of what nitrate will actually look like in real life. notice that resonance is indicated by the dashed lines, meaning that those bonds are neither single or double bonds but a combination of the two. https://en.wikipedia.org/wiki/Nitrate#/media/File:Nitrate-ion-with-partial-charges...
- Tue Nov 06, 2018 8:37 pm
- Forum: Resonance Structures
- Topic: bond lengths of resonant atoms
- Replies: 4
- Views: 572
Re: bond lengths of resonant atoms
when you draw the resonance structures, it seems like the bonds are of different lengths since there's a combination of two single bonds and one double bond (nitrate) or one single bond and one double bond. however, in reality, the bonds in the actual polyatomic ions are neither single or double bon...
- Tue Nov 06, 2018 8:26 pm
- Forum: Bond Lengths & Energies
- Topic: Bond Lengths
- Replies: 5
- Views: 633
Re: Bond Lengths
Also, if you see that your Lewis structure is going to have resonance structures, then you know that all of the bonds involved will have the same length.
- Wed Oct 31, 2018 2:57 pm
- Forum: Resonance Structures
- Topic: Resonance
- Replies: 8
- Views: 978
Re: Resonance
If you're interested, you can search up "ozone lewis structure" and you'll see that it has two resonance structures, but in reality it exists as something in between like this: https://upload.wikimedia.org/wikipedia/ ... dipole.png
- Wed Oct 31, 2018 2:51 pm
- Forum: Resonance Structures
- Topic: resonance hybrids
- Replies: 4
- Views: 513
Re: resonance hybrids
You can write a Lewis structure with dotted lines between the atoms to show resonance, but I think it'd be safer to write out each resonance structure and have the double arrows in between each to show resonance since that's what Dr. Lavelle did in class.
- Wed Oct 31, 2018 2:47 pm
- Forum: Lewis Structures
- Topic: Octet exceptions
- Replies: 2
- Views: 371
Re: Octet exceptions
When the elements that can have an expanded octet form compounds, they are most stable at their lowest energy state. You can determine this by calculating formal charges on each of the elements in the compound. Usually, the Lewis structures with the most 0's/formal charges closest to 0 are the most ...
- Mon Oct 22, 2018 6:32 pm
- Forum: Properties of Electrons
- Topic: Orbitals
- Replies: 5
- Views: 801
Re: Orbitals
3 orbitals are present in the 4p subshell since there are always 3 orbitals in any p subshell - the px, py, and pz.
- Mon Oct 22, 2018 6:30 pm
- Forum: Properties of Electrons
- Topic: Electron Energy
- Replies: 3
- Views: 488
Re: Electron Energy
When an e- transitions from the 1s orbital to the 2p orbital, it is moving to a higher energy state in a different shell (from n=1 to n=2). Since the 2s and 2p orbitals are all in the same shell, they all have the same energy in a hydrogen atom due to its degenerate nature.
- Mon Oct 22, 2018 6:25 pm
- Forum: Properties of Light
- Topic: The Uncertainty Principle
- Replies: 1
- Views: 184
Re: The Uncertainty Principle
1 eV = 1.602 x 10^-19 J
you can find this on the constants and equations link on the class website. it'll also be provided on the constants and equation sheet during tests, so no need to worry about memorizing it!
you can find this on the constants and equations link on the class website. it'll also be provided on the constants and equation sheet during tests, so no need to worry about memorizing it!
- Fri Oct 19, 2018 4:29 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Topic 1E number 13 a)
- Replies: 2
- Views: 81
Re: Topic 1E number 13 a)
Thank you! Does this rule apply to gold also? And in that case, why is the e- configuration for tungsten(W), which is in period 6 under chromium, [Xe]4f^14 5d^4 6s^2 and not [Xe] 4f^14 5d^5 6s^1? (part e) of the same problem)
- Wed Oct 17, 2018 7:29 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Shells, subshells, and orbital
- Replies: 1
- Views: 73
Re: Shells, subshells, and orbital
A subshell and an orbital refer to the same thing, and a shell includes an orbital/subshell. For example, the second shell (n = 2) would have one s orbital and one p orbital.
- Wed Oct 17, 2018 7:26 pm
- Forum: *Shrodinger Equation
- Topic: Schrodinger Equation
- Replies: 2
- Views: 456
Re: Schrodinger Equation
Schrodinger's equation uses wave function to describe an e- in an atom. The Schrodinger equation is H(psi) = E(psi) where psi is the pitchfork looking Greek letter representing the wave function, and what the equation is telling us is that basically after taking the double derivative of the wave fun...
- Wed Oct 17, 2018 7:00 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Topic 1E number 13 a)
- Replies: 2
- Views: 81
Topic 1E number 13 a)
Part a) of number 13 asks for the ground-state electron configuration for silver, and the answer says it's [Kr] 4d^10 5s^1. The textbook explicitly stated that only chromium and copper were exceptions to the general electron configuration rules and didn't include silver but the answer suggests other...
- Tue Oct 09, 2018 10:49 pm
- Forum: Einstein Equation
- Topic: Units
- Replies: 5
- Views: 1033
Re: Units
The unit "hertz" means 1/seconds, so when you multiply J x s by Hz = 1/s, the seconds cancel out and you're left with J.
- Tue Oct 09, 2018 10:47 pm
- Forum: Einstein Equation
- Topic: Units of Hz in Equations
- Replies: 1
- Views: 211
Re: Units of Hz in Equations
MHz means "mega hertz" and is equal to 1,000,000 Hz, so you're right in saying that M is just a prefix. When you do a problem always convert back into the base unit unless the problem specifically tells you otherwise.
- Mon Oct 08, 2018 8:07 pm
- Forum: Photoelectric Effect
- Topic: Rydberg constant
- Replies: 3
- Views: 220
Re: Rydberg constant
There's also a Constants and Equations link on the class website that has the Rydberg constant, Avogadro's number, and a bunch of other constants and useful numbers like the mass of an electron. I'm pretty sure that's the sheet that will be provided to us on tests and exams.
- Mon Oct 08, 2018 8:03 pm
- Forum: Photoelectric Effect
- Topic: Photoelectric Effect
- Replies: 4
- Views: 288
Re: Photoelectric Effect
According to the wave model of light, increasing the intensity of the light should increase the energy of the wave since the amplitude of a wave = its energy and intensity is amplitude squared, so the scientists performing the experiment thought that increasing the intensity of the light should allo...
- Mon Oct 08, 2018 7:50 pm
- Forum: Properties of Light
- Topic: Problem 1A.15. 7th Edition p. 10
- Replies: 1
- Views: 142
Re: Problem 1A.15. 7th Edition p. 10
I had a bit of trouble with this problem as well and ended up just using the guess and check method with Equation 2 on page 7. Since you know that it is part of the Lyman Series, n1 has to = 1 and then I just plugged in values for n2 until the equation equaled itself. For example, I set up the equat...
- Sun Oct 07, 2018 5:27 pm
- Forum: Properties of Light
- Topic: Calculating amount of photons
- Replies: 4
- Views: 655
Re: Calculating amount of photons
The energy per photon can be found using the equation E = hv where h is Planck's constant and v is the frequency, which you can find from the given wavelength with the [speed of light = wavelength x frequency] equation.
- Sun Oct 07, 2018 2:15 pm
- Forum: Properties of Light
- Topic: Calculating amount of photons
- Replies: 4
- Views: 655
Re: Calculating amount of photons
After finding the energy of a photon using the equation, E = hf (f = frequency), you just need to divide the total energy emitted (11J) by the energy per photon to get the number of photons: 11J/(1.074 x 10^-19 J).
- Sun Oct 07, 2018 2:02 pm
- Forum: Properties of Light
- Topic: Chapter 1 Question 1.3
- Replies: 3
- Views: 233
Re: Chapter 1 Question 1.3
You can also look at the graph they provide in Figure 1A.8, which explains why C) is true. Over the same distance, when the wavelength is longer, the frequency is shorter so there is less change overall than short wavelengths with high frequencies.
- Wed Oct 03, 2018 2:42 pm
- Forum: Molarity, Solutions, Dilutions
- Topic: Relationship between density and molarity
- Replies: 2
- Views: 2056
Relationship between density and molarity
Today in discussion we learned that density was related to molarity as they had similar equations (D = m/V and C = n/V) and were given the example: 6.54g (1cm^3/0.7857) = 8.32 cm^3 = 8.32 mL While I understand the calculations and the relationship between cm^3 and mL, I'm not sure where the 6.54 or ...
- Tue Oct 02, 2018 5:14 pm
- Forum: Balancing Chemical Reactions
- Topic: Problem L.35
- Replies: 5
- Views: 359
Re: Problem L.35
I have the 7th edition and had the same problem where I spent forever on the problem because it wouldn't balance. I'm pretty sure it's a typo as well because in the solutions manual they also give the answer by using Fe3Br8.
- Fri Sep 28, 2018 8:49 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Visualizing Orbitals in Atoms
- Replies: 3
- Views: 377
Re: Visualizing Orbitals in Atoms
Thank you, that was a great explanation!
- Fri Sep 28, 2018 8:46 pm
- Forum: Accuracy, Precision, Mole, Other Definitions
- Topic: Precision of One Point [ENDORSED]
- Replies: 2
- Views: 215
Re: Precision of One Point [ENDORSED]
Got it, thanks!
- Fri Sep 28, 2018 5:00 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Visualizing Orbitals in Atoms
- Replies: 3
- Views: 377
Visualizing Orbitals in Atoms
I know we haven't talked about this in class yet, but I was doing some basic chemistry review for a life science class and a problem came up for me about orbitals in atoms. A hydrogen atom is simple enough for visualizing--just a spherical shape for the one s-orbital. But for aluminum, would there b...
- Fri Sep 28, 2018 4:46 pm
- Forum: Accuracy, Precision, Mole, Other Definitions
- Topic: Precision of One Point [ENDORSED]
- Replies: 2
- Views: 215
Precision of One Point [ENDORSED]
Relating back to the visual Dr. Lavelle showed us in lecture today of the points on a target, if I only had one point on the bulls eye, I know that would be accurate but would it also be precise? I'm not sure as to whether just one point can be precise or not.