Search found 75 matches

by BrittneyMyint2C
Sun Jan 17, 2021 5:45 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Textbook Problem 5I23
Replies: 3
Views: 8

Re: Textbook Problem 5I23

Hi! Before the reaction we start off with 0 products. Given from the problem we learn that at equilibrium we have 0.478 mol CH4, which means that 0 + x = 0.478, and x = 0.478. Therefore, the change for CH4 is +0.478 mol. From there we figure out that CO would change by -0.478 mol, H2 changes by -3(0...
by BrittneyMyint2C
Fri Jan 15, 2021 3:58 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: What is autoprotolysis? lavelle's lecture 1/15 [ENDORSED]
Replies: 7
Views: 49

Re: What is autoprotolysis? lavelle's lecture 1/15 [ENDORSED]

As the comment above says, autoprotolysis is a proton transfer between two like molecules, so in this case the proton transfer between two H2O molecules.
by BrittneyMyint2C
Wed Jan 13, 2021 11:45 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: pH and pKa
Replies: 6
Views: 26

Re: pH and pKa

Hi, I believe that pH would be more used for showing the concentration of H3O+ since pH is gotten from -log[H3O+] while pKa is more related to dissociation of the acid since it's gotten from -log(Ka). It may also depend on the information given and what the question is asking. Hope this helps!
by BrittneyMyint2C
Wed Jan 13, 2021 10:18 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Ka and Weak Acids
Replies: 3
Views: 21

Re: Ka and Weak Acids

Hello, if Ka is large, then there are more products than reactants, so the acid dissociates and is therefore strong. If Ka is small, then there are more reactants than products, so the acid would be weak since it does not completely dissociate. For example, if we didn't know that HCl was a strong ac...
by BrittneyMyint2C
Wed Jan 13, 2021 10:03 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 5I.13 Part C
Replies: 2
Views: 12

Re: 5I.13 Part C

Hi, for part c, you determine whether F2 or Cl2 is more thermodynamically stable by looking at the K values. In general, if K has a relatively large value, this means that the products are more thermodynamically stable since the products are more favored; if K has a smaller value, then the reactants...
by BrittneyMyint2C
Tue Jan 12, 2021 8:15 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Textbook Focus Problem 5.33 a
Replies: 3
Views: 29

Re: Textbook Focus Problem 5.33 a

As the above post says, breaking a bond requires energy, so this kind of reaction is endothermic. In contrast, forming a bond to make a more stable molecule releases energy, so this kind of reaction is exothermic. Hope this helps!
by BrittneyMyint2C
Sun Jan 10, 2021 4:55 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Textbook problem 5J.5
Replies: 3
Views: 26

Re: Textbook problem 5J.5

Pressure and Volume are inversely related. The smaller the volume, the higher the pressure. When the total pressure is increased, the side with less moles is favored. For example, part (a) is 2 moles of gas on the reactants and 3 moles of gas on the products. Since there is a greater amount of mole...
by BrittneyMyint2C
Thu Jan 07, 2021 10:36 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Audio visual module pt 4
Replies: 3
Views: 29

Re: Audio visual module pt 4

When the problem said adding water, it was referring to increasing the concentration of H2O on the reactants side. Doing this would make Q<K at that moment, so to reestablish equilibirum, the reaction would proceed to the right and make more product, therefore increasing product concentration. And y...
by BrittneyMyint2C
Thu Jan 07, 2021 2:43 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Sapling #5 Multistep Eqm Constants
Replies: 3
Views: 31

Re: Sapling #5 Multistep Eqm Constants

Hi, for that question you are basically "adding" the listed reactions to get the desired reaction. I would first try finding the equations that have the species that are in the desired reaction and then try to arrange the reactions by using the reverse reaction and multiplying to get close...
by BrittneyMyint2C
Thu Jan 07, 2021 2:19 am
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Effects of Decreasing Temperature on reactions
Replies: 4
Views: 33

Re: Effects of Decreasing Temperature on reactions

I agree with the post above. Decreasing temperature for an endothermic reaction would favor the reverse reaction (equilibrium shifts left) while decreasing the temperature for an exothermic reaction would favor the forward reaction (equilibrium shifts right).
by BrittneyMyint2C
Wed Jan 06, 2021 4:09 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Lecture 2 Ex) 1.5mol PCl5
Replies: 2
Views: 21

Re: Lecture 2 Ex) 1.5mol PCl5

Hi, I believe for this question you should be trying to find the concentrations of each substance at equilibrium using the given K. You would first calculate the initial molar concentration of PCl5: 1.50 mol/500mL = 3.00 M. Write a balanced equation for the reaction and then use an ICE table. Initia...
by BrittneyMyint2C
Mon Jan 04, 2021 2:02 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Chemical Equilibrium part 2 post assessment #24
Replies: 1
Views: 20

Re: Chemical Equilibrium part 2 post assessment #24

Hi! To determine if SO3 will be composed or formed, you should compare Q and K. If Q<K, then the forward reaction is favored, so more product will be formed until the reaction is at equilibrium. If Q>K, then the reverse reaction is favored, so the product will decompose to make more reactants until ...
by BrittneyMyint2C
Fri Dec 11, 2020 9:13 pm
Forum: Properties & Structures of Inorganic & Organic Acids
Topic: Strength of Acids
Replies: 8
Views: 64

Re: Strength of Acids

From my understanding, the H in each of those acids (HClO, HBrO, and HIO) are bonded to the O, so you have to consider the electronegativity rather than the size. However, if it was HCl, HBr, and HI, since the H are bonded to the anions, you can consider the size. Hope this helps!
by BrittneyMyint2C
Fri Dec 11, 2020 1:32 pm
Forum: Lewis Structures
Topic: 2C.1
Replies: 1
Views: 26

Re: 2C.1

Looking through the other posts regarding this question, I believe there is a mistake in the online answer key. The radicals should be CH3 and OH. Hope this helps!
by BrittneyMyint2C
Thu Dec 10, 2020 9:02 pm
Forum: Identifying Acidic & Basic Salts
Topic: Textbook Problem J.17
Replies: 2
Views: 37

Re: Textbook Problem J.17

I believe Dr. Lavelle said that cations from Group 1 and 2 are weak Lewis acids and don't generate H3O+, so that means that for parts a and b, we would use the anions (since Na+ in part a and K+ in part b wouldn't generate H3O+). I used the cations for c and d because Dr. Lavelle had also said that ...
by BrittneyMyint2C
Thu Dec 10, 2020 8:27 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: Sites of Hydrogen Bonding (Review Powerpoint Slide 9)
Replies: 2
Views: 36

Re: Sites of Hydrogen Bonding (Review Powerpoint Slide 9)

Hi, so I'm also a little iffy on hydrogen bonding sites, but hydrogen bonds are formed between a hydrogen atom bonded to a strongly electronegative atom (like N, O, and F) of one molecule and a N, O, or F of another molecule. The lone pairs of these electronegative atoms form hydrogen bonds. So for ...
by BrittneyMyint2C
Thu Dec 10, 2020 7:41 pm
Forum: Conjugate Acids & Bases
Topic: Sapling HW 10 Question 2
Replies: 2
Views: 26

Re: Sapling HW 10 Question 2

Hi, so conjugate base of an acid is the acid after it has donated an H+. For HSO4-, after it donates its H+, it would be SO4 2-. So, the conjugate base for HSO4- is SO4 2-. A conjugate acid of a base is the base after it gains an H+. For example, if HSO4- was a base, then its conjugate acid would be...
by BrittneyMyint2C
Sat Dec 05, 2020 1:07 pm
Forum: Conjugate Acids & Bases
Topic: 6A.1 Question
Replies: 1
Views: 42

Re: 6A.1 Question

Hi! So a conjugate base of an acid is the result of the species after it donates a H+, while a conjugate acid of a base is the result of a species after it accepts a H+. For example, the conjugate acid of CH3NH2 would be CH3NH2 after it accepts a H+, so the result would be CH3NH3+. The conjugate bas...
by BrittneyMyint2C
Sat Dec 05, 2020 12:59 pm
Forum: Properties & Structures of Inorganic & Organic Acids
Topic: 6A 1a
Replies: 2
Views: 29

Re: 6A 1a

If you try drawing the Lewis structure for CH3NH2, you see that C already has a full octet since it is bonded to 3 H atoms and 1 N atom. So, when it gains a proton, the H+ cannot bond to C. Instead, the proton bonds to the N atom to make CH3NH3+ because N has a lone pair. Hope this makes sense!
by BrittneyMyint2C
Fri Dec 04, 2020 6:35 pm
Forum: Hybridization
Topic: 2.57 textbook
Replies: 1
Views: 34

Re: 2.57 textbook

Hi. I found a response that was posted a while ago, hope this helps:
viewtopic.php?f=42&t=68618&p=276278&sid=15282129c8445a7ac8c776092c3491af#p276278
by BrittneyMyint2C
Fri Dec 04, 2020 12:30 am
Forum: Bronsted Acids & Bases
Topic: Calcium Oxide Example from 12/2 Lecture
Replies: 1
Views: 28

Re: Calcium Oxide Example from 12/2 Lecture

Hi, CaO dissociates into Ca2+ and O2- because CaO is a strong base. The oxygen from the CaO accepts the proton (H+) from water, which in turn makes one of the H-O bonds from the water break since H can only hold 2 electrons. So, we have 2 OH- as a result as well as a Ca2+ ion. Hope this makes sense!
by BrittneyMyint2C
Fri Dec 04, 2020 12:22 am
Forum: Bronsted Acids & Bases
Topic: Question about net ionic equation for neutralization of weak acids/bases
Replies: 2
Views: 31

Re: Question about net ionic equation for neutralization of weak acids/bases

Hello, I believe it means that when we're writing a net ionic equation and there's a weak acid, we leave it in the molecule's formula. So if the acid is HF for example, we leave it as HF instead of writing it as H+ and F- because HF is a weak acid. On the other hand, when there's a strong acid, such...
by BrittneyMyint2C
Wed Dec 02, 2020 12:33 pm
Forum: Properties & Structures of Inorganic & Organic Acids
Topic: Organic Acids
Replies: 2
Views: 28

Re: Organic Acids

Yes, strong acids (almost) completely dissociate in water. Since organic acids do not completely dissociate in water, they're weaker. Hope this helps!
by BrittneyMyint2C
Sat Nov 28, 2020 3:25 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Axial vs Equatorial
Replies: 4
Views: 46

Re: Axial vs Equatorial

Hello, axial atoms are vertically away from each other, so they form a 180 degree angle. Equatorial atoms are along the same equatorial plane. For example, in a trigonal bipyramidal shape, 2 atoms are axial and 3 are equatorial. I usually try to visualize the diagrams of the shapes of the molecules ...
by BrittneyMyint2C
Sat Nov 28, 2020 3:18 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Textbook 2E #25 part a
Replies: 3
Views: 24

Re: Textbook 2E #25 part a

If you visualize a tetrahedral molecule you see that regardless of where you put the Cl or H, the Cl and Cl will always be next to each other, so CH2Cl2 will always be polar. For Lewis structures, I believe it doesn't really matter if you put the H or Cl next to each other or not. Hope this helps!
by BrittneyMyint2C
Sat Nov 28, 2020 3:07 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Textbook 2E #19 part b
Replies: 1
Views: 23

Re: Textbook 2E #19 part b

Hi, when it says that the shape is tetrahedral it is referring to the shape around the carbon atoms. From your description I believe your initial Lewis structure was correct (with 3 hydrogens around each carbon and each carbon being bonded to the Be atom), so if you look at your Lewis structure you ...
by BrittneyMyint2C
Sat Nov 28, 2020 2:58 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Textbook Question 2E.29
Replies: 1
Views: 25

Re: Textbook Question 2E.29

Hi, yes you're correct that it has to do with the direction of the dipoles. 3 doesn't have a dipole moment since its dipoles cancel out. The way I thought of determining whether 1 or 2 has the larger dipole moment was that the vectors that are pointing in a similar direction would be stronger, while...
by BrittneyMyint2C
Thu Nov 26, 2020 8:28 pm
Forum: *Molecular Orbital Theory (Bond Order, Diamagnetism, Paramagnetism)
Topic: Ligands
Replies: 9
Views: 102

Re: Ligands

Since ligands are Lewis bases (electron pair donors), I believe you would try identifying which molecule is donating an electron pair to form a coordinate covalent bond. From there, you could possibly determine the number of ligands. As a result of coordinate covalent bond formation, a coordination ...
by BrittneyMyint2C
Thu Nov 26, 2020 8:20 pm
Forum: Hybridization
Topic: Sapling #12
Replies: 27
Views: 172

Re: Sapling #12

Hello, as the person above said, the hybridization of the atoms depend on the region of electron density. From the Lewis structure, you see that both C and O have four regions of electron density, so the hybridization for the oxygen and carbon is sp3 hybridization. Hope this helps!
by BrittneyMyint2C
Wed Nov 25, 2020 3:10 pm
Forum: Hybridization
Topic: Ethylene Example
Replies: 1
Views: 27

Re: Ethylene Example

I think it is because of the differences in the number of electron density, since NH3 has 4 regions of electron density while C2H4 has 3 regions of electron density. Also, the "lone electron" goes to the 2p orbital rather than sp2 because of electron repulsion and there being a smaller ene...
by BrittneyMyint2C
Mon Nov 23, 2020 1:04 pm
Forum: Hybridization
Topic: Hybrid Bonds
Replies: 3
Views: 41

Re: Hybrid Bonds

Follow up question, when do we know a molecule uses hybrid orbitals then?
by BrittneyMyint2C
Sun Nov 22, 2020 1:19 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Tetrahedral vs. Square Planar
Replies: 4
Views: 39

Re: Tetrahedral vs. Square Planar

Hi, the difference between the two shapes is if the central atom has lone pairs. If a molecule is tetrahedral, then the central atom is bonded to 4 atoms and has no lone pairs , but if it is square planar, then the central atom is bonded to 4 atoms and has 2 lone pairs . Drawing a lewis structure ma...
by BrittneyMyint2C
Sat Nov 21, 2020 1:33 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Which shapes to memorize
Replies: 11
Views: 133

Re: Which shapes to memorize

I'm assuming it's all the shapes that Dr. Lavelle mentioned in the lectures: - linear : central atom has 2 bonding pairs; 180 degree bond angle - trigonal planar: central atom has 3 bonding pairs, 120 degree bond angles - tetrahedral: central atom has 4 bonding pairs, 109.5 degree bond angle - trigo...
by BrittneyMyint2C
Fri Nov 20, 2020 7:00 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: determining molecular shape
Replies: 10
Views: 78

Re: determining molecular shape

Hi, as everyone has said, the Lewis structures of CO2 and SO2 show that the central atom in CO2 does not have a lone pair while the central atom of SO2 does. So, the lone pair on the S atom causes lone pair and bonding pair repulsion, making SO2 have a bent shape. If you draw the vectors, you find t...
by BrittneyMyint2C
Mon Nov 16, 2020 12:35 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Trigonal Pyramidal Shape in the Sulfite Ion
Replies: 4
Views: 38

Re: Trigonal Pyramidal Shape in the Sulfite Ion

Yes you're correct! The carbonate ion has 3 regions of electron density while the sulfite ion has 4 regions of electron density (with the central atom having 3 bonding pairs and 1 lone pair), so the carbonate ion has a trigonal planar shape while the sulfite ion has a trigonal pyramidal shape.
by BrittneyMyint2C
Mon Nov 16, 2020 12:30 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: SO3 Structure
Replies: 4
Views: 45

Re: SO3 Structure

Hi, as others have said, one of the bonding oxygen atoms would make a double bond, so the sulfur would still have a lone pair. Therefore, even with the double bond, 3 of the 4 bonding positions are occupied by atoms, so the shape is trigonal pyramidal.
by BrittneyMyint2C
Sat Nov 14, 2020 6:03 pm
Forum: Lewis Structures
Topic: Textbook problem 2C3 part c
Replies: 1
Views: 29

Textbook problem 2C3 part c

For 2C3, it asks us to "Draw the Lewis structure, including typical contributions to the resonance structure (where appropriate, allow for the possibility of octet expansion, including double bonds in different positions), for (a) periodate ion; (b) hydrogen phosphate ion; (c) chloric acid; (d)...
by BrittneyMyint2C
Sat Nov 14, 2020 3:19 pm
Forum: Dipole Moments
Topic: Strength of intermolecular forces
Replies: 3
Views: 46

Re: Strength of intermolecular forces

I'm a bit confused on the last half of the sentence,, but I believe what the first half is trying to say is that as the distance increases between the molecules, the strength of the dipole-dipole interaction will decrease. Hope this helps!
by BrittneyMyint2C
Sat Nov 14, 2020 1:44 am
Forum: Quantum Numbers and The H-Atom
Topic: Textbook Problems 1D.1 and 1E.1
Replies: 2
Views: 36

Re: Textbook Problems 1D.1 and 1E.1

Hi, so for 1D1, we are going from 1s to 2p. a) The energy of the electron will increase because we are moving to a higher energy level. b) Because we're going from n =1 to n=2, we see that the value of n increases. c) The value of l increases because l = 0 is s and l = 1 is p. d) Since we're moving ...
by BrittneyMyint2C
Sat Nov 14, 2020 1:17 am
Forum: Ionic & Covalent Bonds
Topic: London Dispersion Forces
Replies: 11
Views: 71

Re: London Dispersion Forces

Hello! Some main things you should know about London dispersion forces is that it is a temporary intermolecular force that occurs in all molecules. So, while the other forces tend to have molecules that are polar, like in dipole-dipole, London dispersion forces can occur even between nonpolar molecu...
by BrittneyMyint2C
Wed Nov 11, 2020 7:19 pm
Forum: Lewis Structures
Topic: homework 2B.3
Replies: 5
Views: 44

Re: homework 2B.3

BrF3 has a total of 28 electrons (7 electrons from Br, 7 electrons from each F = 7 +3(7) = 28). When you create the Lewis structure, we make single bonds between Br and each of the F atoms, so we used 6 electrons. After we put the lone pairs for each F, we now have 4 electrons left. So, we put these...
by BrittneyMyint2C
Tue Nov 10, 2020 4:20 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Week 6 Discussion Question- Cobalt
Replies: 1
Views: 16

Re: Week 6 Discussion Question- Cobalt

Because it gives you the electron configuration, you should base your answer off of that. The electron configuration for M^2+ is [Ar]3d^7, which means that the ground state electron configuration is [Ar]3d^74s^2, because the electrons are first removed from the 4s shell. If you look at the periodic ...
by BrittneyMyint2C
Mon Nov 09, 2020 12:54 pm
Forum: Dipole Moments
Topic: Dipole clarification
Replies: 6
Views: 31

Re: Dipole clarification

I would say that dipole moments apply more to covalent bonds, since we're considering whether a covalent molecule is polar or nonpolar due to electronegativity differences.
by BrittneyMyint2C
Sun Nov 08, 2020 11:33 pm
Forum: Ionic & Covalent Bonds
Topic: HW Question
Replies: 4
Views: 54

Re: HW Question

The molecule would have London dispersion forces and dipole dipole. Dispersion forces are in all molecules. Since there is an oxygen in the molecule with C and H, this makes the molecule polar, so there would be dipole dipole intermolecular forces. Hydrogen forces would not be present because the hy...
by BrittneyMyint2C
Fri Nov 06, 2020 3:31 pm
Forum: Octet Exceptions
Topic: Boron Trifluoride Example
Replies: 2
Views: 42

Re: Boron Trifluoride Example

Hi, we cannot use double bonds for BF3 because if there is a double bond between B and F, the formal charge of F would change from 0 to +1 and the formal charge of B would change from 0 to -1. Since F has a high electron affinity, this formal charge wouldn't make sense. Hope this helps!
by BrittneyMyint2C
Fri Nov 06, 2020 1:35 pm
Forum: Lewis Structures
Topic: Textbook 2A. 23
Replies: 5
Views: 52

Re: Textbook 2A. 23

I don't think memorization is required, the periodic table is helpful. To follow on the previous response, B) indium(III) sulfide, Indium is +3 and sulfide -2, so to balance the charges, you will have In2S3 Hi! I was wondering how would I find the charge of the element on the periodic table? Hello,...
by BrittneyMyint2C
Thu Nov 05, 2020 2:45 pm
Forum: Lewis Structures
Topic: Sapling HW Question 3
Replies: 5
Views: 52

Re: Sapling HW Question 3

Hi! I have another question with this problem. Can someone explain what the difference is between the endings "-ite" and "-ate"? What do they tell us about the ion/chemical formula? Hello! I believe -ate is used when an ion has one more oxygen atom than those with -ite endings. ...
by BrittneyMyint2C
Wed Nov 04, 2020 12:52 pm
Forum: Octet Exceptions
Topic: Are all octet exceptions more reactive?
Replies: 5
Views: 63

Re: Are all octet exceptions more reactive?

I believe this only applies to radicals since radicals have an unpaired valence electron, which makes it more reactive. Hope this helps!
by BrittneyMyint2C
Wed Nov 04, 2020 12:45 pm
Forum: Octet Exceptions
Topic: PCl5 octet exception
Replies: 1
Views: 33

Re: PCl5 octet exception

Hi, if you look at the PCl5 Lewis structure, you see that P forms a bond with the each of the 5 fluorines, so it has 10 valence electrons. Since an element with an octet would have s2p6, P has s2p6d2 because it has an expanded octet. It uses the d orbital to accommodate the additional 2 electrons. A...
by BrittneyMyint2C
Sun Nov 01, 2020 4:53 pm
Forum: Einstein Equation
Topic: homework weeks 2,3,4 #12
Replies: 4
Views: 62

Re: homework weeks 2,3,4 #12

Hi, so for the first part, you don't have to multiply 1.028 by Avogadro's constant, so the answer is just 1.028 eV/atom. For the second part, the conversion from eV to J is 1.602*10^-19 J, so you may have forgotten the negative sign. I would recommend just using the answer you got from the first par...
by BrittneyMyint2C
Sat Oct 31, 2020 11:13 pm
Forum: Ionic & Covalent Bonds
Topic: Ionic Vs. Covalent
Replies: 7
Views: 63

Re: Ionic Vs. Covalent

Hi, as everyone as said, ionic bonds are formed through the transfer of electrons. Metals tend to lose the electrons (and become cations) while nonmetals gain the electrons (and become anions). An example of this would be NaCl. Covalent bonds are formed through the sharing of electrons, which are be...
by BrittneyMyint2C
Wed Oct 28, 2020 8:34 pm
Forum: Trends in The Periodic Table
Topic: Size of isoelectronic atoms/ions
Replies: 9
Views: 75

Re: Size of isoelectronic atoms/ions

lwong Dis1L wrote:How do you know if an atom is isoelectronic in the first place?

Atoms that are isoelectronic would have the same electron configuration. For example, F- and Ne have the same electron configuration, 1s^2 2s^2 2p^6, since F gained an electron. Hope this helps!
by BrittneyMyint2C
Wed Oct 28, 2020 8:21 pm
Forum: Photoelectric Effect
Topic: Photoelectric Effect post module
Replies: 4
Views: 60

Re: Photoelectric Effect post module

Hello! 3.607 x 10-19 J will be the threshold energy, since it is this energy is required to remove an electron. Since there's 0 kinetic energy, this means that E(photon) = threshold energy, based on the E(photon) - threshold energy = kinetic energy equation. So if we want to see the longest waveleng...
by BrittneyMyint2C
Wed Oct 28, 2020 12:14 am
Forum: Quantum Numbers and The H-Atom
Topic: Quantum Numbers
Replies: 3
Views: 42

Re: Quantum Numbers

Hi, I'm a bit confused on the wording of the question, but I'll try answering from both of my interpretations: If you mean "can the quantum numbers could be the same, like n =1, l = 1, ml = 1", then no, the first three quantum numbers cannot be the same since the allowed values of l is fou...
by BrittneyMyint2C
Mon Oct 26, 2020 5:03 pm
Forum: Properties of Light
Topic: Practice Question #6: The work function for lithium is 4.6x10^-19 J.
Replies: 2
Views: 39

Re: Practice Question #6: The work function for lithium is 4.6x10^-19 J.

Part b is basically asking for the kinetic energy, so you'll need the E(photon) - work function = kinetic energy later. First convert 7.3x10^14 Hz to energy with E = hv equation to get E(photon). Then subtract this with 4.6x10^-19 J (work function) to get the kinetic energy.
Hope this helps!
by BrittneyMyint2C
Sat Oct 24, 2020 3:13 pm
Forum: Properties of Electrons
Topic: Sapling #23
Replies: 4
Views: 41

Re: Sapling #23

They told us "the electron affinity is the difference in energy between the incident photons and the energy of the ejected electrons", so that means we need the energy of the incident light first. I first converted the incident light wavelength into energy using the E = hc/λ equation, and ...
by BrittneyMyint2C
Thu Oct 22, 2020 6:10 pm
Forum: Einstein Equation
Topic: Sapling Number 4
Replies: 6
Views: 92

Re: Sapling Number 4

So our main goal is to get the number of electrons, and we can find this once we get the number of photons since there's a one photon one electron interaction, so number of photons = number of electrons ejected. We have to divide because the threshold energy we calculated only gives us the energy pe...
by BrittneyMyint2C
Thu Oct 22, 2020 5:52 pm
Forum: *Shrodinger Equation
Topic: Schrodinger Equation
Replies: 3
Views: 57

Schrodinger Equation

Hi, what does Dr. Lavelle mean when he says the wave function "works" or when HΨ is a "valid model"? Does it depend on the energy of the electron we get from the HΨ = EΨ equation?
Thanks in advance!
by BrittneyMyint2C
Thu Oct 22, 2020 4:53 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Hydrogen Emission Spectrum
Replies: 3
Views: 46

Re: Hydrogen Emission Spectrum

Hi! So since they give you the frequency, you can use the c=hλ equation to figure out the wavelength. By doing this, we find that the wavelength is about 486 nm, so this is in the visible light region. Since the experiment takes place in hydrogen gas, we know that the visible light region is related...
by BrittneyMyint2C
Tue Oct 20, 2020 11:11 pm
Forum: Einstein Equation
Topic: Sapling Number 4
Replies: 6
Views: 92

Re: Sapling Number 4

For the first question, your main goal is to use the E(photon) - E(kinetic energy) = work function to ultimately determine the work function/threshold energy. the frequency given will allow you to determine the E(photon) using the E=hv formula. The maximum kinetic energy will be the E(kinetic energy...
by BrittneyMyint2C
Tue Oct 20, 2020 10:18 pm
Forum: Properties of Light
Topic: Relationship between intensity and amplitude
Replies: 3
Views: 47

Re: Relationship between intensity and amplitude

Hello! As said above, I believe the professor did mention in lecture or an audio/visual topic (the photoelectric effect one, if you would like to review it) that increasing the amplitude increases the intensity of light.
by BrittneyMyint2C
Sun Oct 18, 2020 7:00 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Heisenberg Uncertainty Post Module Q14
Replies: 1
Views: 20

Re: Heisenberg Uncertainty Post Module Q14

I believe the answer is D which is "No, the uncertainties in position, speed, and momentum of a moving object are not noticeable or measurable." because Heisenberg uncertainty principle is concerned with moving objects. Hope this helps!
by BrittneyMyint2C
Thu Oct 15, 2020 11:15 pm
Forum: Photoelectric Effect
Topic: Combining Equations
Replies: 4
Views: 52

Re: Combining Equations

Hi! So basically what he did was he found a variable that was in both equations, which would be v (frequency). So to get the E = hc/λ formula, he used the c=λv formula and isolated v in this equation, so v=c/λ. When you substitute this into the E=hv formula, you replace v with c/λ, so you get E=h(c/...
by BrittneyMyint2C
Wed Oct 14, 2020 8:41 pm
Forum: Photoelectric Effect
Topic: Excess energy
Replies: 20
Views: 135

Re: Excess energy

LaurenChoi_1J wrote:Ahh got it! In the lecture, when he said that e(excess)=1/2mv^2, does that mean the amount of excess energy equal to the kinetic energy?

Hi! Yes, I believe he said that when there is excess energy, the excess energy is the kinetic energy. Hope this helps!
by BrittneyMyint2C
Wed Oct 14, 2020 8:38 pm
Forum: Photoelectric Effect
Topic: Electron not emitted even for high intensity light
Replies: 9
Views: 92

Re: Electron not emitted even for high intensity light

Hi! I believe you are talking about how electrons are not emitted, even for high intensity light, when the light has low frequency (long wavelength). This is because all the photons in this wavelength will have the same amount of energy. If the photons in the wavelength do not have sufficient energy...
by BrittneyMyint2C
Wed Oct 14, 2020 8:04 pm
Forum: Properties of Light
Topic: Exercise 1A.15
Replies: 2
Views: 49

Re: Exercise 1A.15

Hello! The question asks for the initial and final energy levels, which means that we eventually need to use the equation v = -R[(1/n1^2) - (1/n2^2)], with v = frequency. Something to note is that since the wavelength was 102.6 nm, this is in the UV series, so the final energy level should be n=1 (w...
by BrittneyMyint2C
Tue Oct 13, 2020 2:31 pm
Forum: Properties of Light
Topic: video modules
Replies: 10
Views: 95

Re: video modules

They are not mandatory, but I did find them helpful! I believe Professor Lavelle went over wave properties of light in his friday lecture and he kinda touched upon the other video modules in yesterday's lecture, but it's good review :)
by BrittneyMyint2C
Mon Oct 12, 2020 12:54 pm
Forum: Properties of Light
Topic: Photoelectric Effect Post-Module Assessment #33
Replies: 4
Views: 45

Re: Photoelectric Effect Post-Module Assessment #33

Does anyone know how to know whether to use use the Ephoton equation versus the other energy/wavelength equations? Hi, I would say it depends on the given information. I would try to figure out the knowns and unknowns, and then looking at the available equations, you could find which equations you ...
by BrittneyMyint2C
Sat Oct 10, 2020 5:45 pm
Forum: Photoelectric Effect
Topic: Question 30 on Module
Replies: 1
Views: 25

Re: Question 30 on Module

Hi! Yes, for this question, you would calculate E(photon) using the equation E(photon) - work function = kinetic energy to find the frequency of the light. So you would use the values you calculated you calculated in the previous questions to find E(photon), and then use the E = hv equation to find ...
by BrittneyMyint2C
Wed Oct 07, 2020 9:59 pm
Forum: Limiting Reactant Calculations
Topic: Dis 1L Week 1 WS #6
Replies: 4
Views: 44

Re: Dis 1L Week 1 WS #6

Hi! To find how much excess reactant was used in the reaction, you would first need to see how much of the excess reactant was consumed during the reaction. To do this, use stoichiometry/mole-to-mole ratios between the limiting reactant (O2) and excess reactant (glucose). 0.156 mol O2 * (1 mol gluco...
by BrittneyMyint2C
Wed Oct 07, 2020 11:46 am
Forum: Limiting Reactant Calculations
Topic: Discussion worksheet question
Replies: 3
Views: 36

Re: Discussion worksheet question

Hi! Percent yield is (actual yield/theoretical yield) * 100%. In experiments, it is unlikely to get a 100% percent yield since there tends to be factors unaccounted for (like side reactions, impurities etc.), leading to the actual yield being less than the theoretical yield. If we were to think of t...
by BrittneyMyint2C
Tue Oct 06, 2020 1:32 am
Forum: Accuracy, Precision, Mole, Other Definitions
Topic: Formula Units vs Molecules/Atoms
Replies: 3
Views: 45

Re: Formula Units vs Molecules/Atoms

In section E.2 of the textbook, I believe it says that formula units is used for ionic compounds, so that may be what differentiates it?
by BrittneyMyint2C
Tue Oct 06, 2020 1:26 am
Forum: Molarity, Solutions, Dilutions
Topic: G5
Replies: 3
Views: 74

Re: G5

I would say that c is pretty similar to parts a and b conceptually, excluding the mole-to-mole ratios we used for those two parts. You would convert 50.0 mg Na2CO3 to moles using its molar mass, and once you get this value (4.72*10^-4 mol), you would use it in the molarity equation, M = n/V. Using t...
by BrittneyMyint2C
Mon Oct 05, 2020 11:55 am
Forum: Balancing Chemical Reactions
Topic: Balancing chemical equations involving combustion
Replies: 15
Views: 117

Re: Balancing chemical equations involving combustion

Yes, O2 is a reactant and combustion of hydrocarbons also produces CO2 and H2O as well!
by BrittneyMyint2C
Sat Oct 03, 2020 11:26 pm
Forum: Molarity, Solutions, Dilutions
Topic: G.5 [ENDORSED]
Replies: 5
Views: 97

Re: G.5 [ENDORSED]

I took a similar approach to Sami. First we determine the moles of Na2CO3 and use this to find the molarity, which is about 0.0796 mol/L. Now, since the mole-to-mole ratio between Na+ and Na2CO3 is 2:1 (2 mol Na+ for every 1 mol Na2CO3), this suggests that there would be 0.00215 mol Na+ for every 0....

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