Search found 81 matches

by Justin Vayakone 1C
Tue Jan 28, 2020 10:40 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 4E.5, 4E.7
Replies: 3
Views: 9

Re: 4E.5, 4E.7

For 4E.5) you used the C-C bond enthalpy (-348kJ/mol) when it should be the C-H bond enthalpy which is 518kJ/mol For 4E.7) a) Your work looks right. You should be getting -202 kJ/mol. Maybe you plugged it into your calculator wrong? b) You forgot to include the bond enthalpy of C-C in the products. ...
by Justin Vayakone 1C
Tue Jan 28, 2020 10:15 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: 4A.9
Replies: 1
Views: 11

Re: 4A.9

Heat released = - heat absorbed
Your work here looks right. I'm assuming you used the negative to change (T-100) into (100-T) right?
by Justin Vayakone 1C
Tue Jan 28, 2020 9:20 pm
Forum: Calculating Work of Expansion
Topic: 4A.3
Replies: 2
Views: 14

Re: 4A.3

My answer book shows 28J. Your statement regarding internal energy and work is correct.
by Justin Vayakone 1C
Tue Jan 28, 2020 9:08 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 4D.9
Replies: 1
Views: 13

Re: 4D.9

I think enthalpy density is a value like regular density or volume, in which the value can't be negative.
by Justin Vayakone 1C
Tue Jan 21, 2020 12:18 am
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: 5.35 Textbook
Replies: 2
Views: 29

Re: 5.35 Textbook

Well let's look at the changes for A B and C. A: -10 B: +5 C: +10 The ratio of A/B/C is -10/+5/+10 or simplified -2/1/2. Negative change indicates the reactant while positive indicates products. The ratio above represents the coefficients for the equation. For every 2 A molecules used, 1 B and 2 C m...
by Justin Vayakone 1C
Mon Jan 20, 2020 11:36 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 6A.23
Replies: 3
Views: 20

Re: 6A.23

Ba(OH)2 is a strong acid, so you can assume all of the reactant turns into products. No need for an ICE table or K value. Just be careful and know that there will be double [OH] compared to [Ba2+].
by Justin Vayakone 1C
Mon Jan 20, 2020 11:32 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: 6C.7
Replies: 1
Views: 29

Re: 6C.7

NH3OH+ is the conjugate acid of hydroxylamine (NH2OH) which is given in Table 6C.2. (CH3)2NH2+ is the conjugate acid of dimethylamine, (CH3)2NH which is also given in Table 6C.2. Just do 14 - pKb to find pkA.
by Justin Vayakone 1C
Mon Jan 20, 2020 11:19 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Kc to Kp conversion
Replies: 2
Views: 17

Re: Kc to Kp conversion

That equation given by your TA comes from this equation (PV=CRT) given by Lavelle. Here is an example I made illustrating where the difference in moles comes from.
Edit: accidentally put exponent inside B brackets twice
by Justin Vayakone 1C
Mon Jan 20, 2020 10:56 pm
Forum: *Making Buffers & Calculating Buffer pH (Henderson-Hasselbalch Equation)
Topic: 6G.5
Replies: 1
Views: 17

Re: 6G.5

NaCl doesn't affect the pH of the solution because the Na+ and Cl- ions don't react with anything. This means we completely ignore NaCl. We are given 0.20 M of NH2NH2, which is a weak base. Create an ICE table and look up the Kb for NH2NH2 in one of the tables in the textbook (6C.1 , 6C.2 and 6E.1)....
by Justin Vayakone 1C
Wed Jan 15, 2020 3:18 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Homework Question 5I.13
Replies: 2
Views: 31

Re: Homework Question 5I.13

When doing the approximation, you have to double check the ionization percentage. Lavelle told us in lecture today that if this ionization percentage is above 5%, then the approximation isn't accurate enough, and we should use other methods like quadratic formula. To find this percentage, do the amo...
by Justin Vayakone 1C
Wed Jan 15, 2020 2:58 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 6B.9a
Replies: 1
Views: 12

Re: 6B.9a

Yeah it looks like the answer key is wrong. The pH should be -0.176. With this pH, the pOH would now be 14.176 and =
by Justin Vayakone 1C
Wed Jan 15, 2020 2:48 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 6B.5
Replies: 1
Views: 14

Re: 6B.5

For each mole of Ba(OH)2 that dissociates, 2 moles of OH are created, so 0.0092 M Ba(OH)2 turns into 0.0184 M OH.
by Justin Vayakone 1C
Tue Jan 14, 2020 11:42 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 5H.2
Replies: 2
Views: 25

Re: 5H.2

My solutions manual only has odds and not even numbered problems. Could you post a picture of it so I could see?
by Justin Vayakone 1C
Tue Jan 14, 2020 11:24 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Thermodynamically Stable?
Replies: 3
Views: 25

Re: Thermodynamically Stable?

Being thermodynamically stable means being in a low energy state. The less energy or enthalpy a molecule has, the more stable it is. Now to the problem, we are asked which reactant (Cl2 or F2) is more stable. The more stable a reactant is, the less it wants to dissociate into product. Comparing 1.1x...
by Justin Vayakone 1C
Mon Jan 13, 2020 1:46 pm
Forum: Properties of Electrons
Topic: Experiments of electrons
Replies: 1
Views: 34

Re: Experiments of electrons

The photoelectric experiment (photons shined on a metal surface which ejected electrons) showcases the particle property of electrons. For the wave property of electrons, I don't know the name, but I think the experiment involved particles passing through a hole in a wall, then a detector at the end...
by Justin Vayakone 1C
Tue Jan 07, 2020 5:54 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 5.35 Part b
Replies: 2
Views: 24

5.35 Part b

This is the answer to part b: (in attachment). I understand the arrangement of the equilibrium constant and where the values (5, 10, and 18) come from, but I don't understand why these values are over 100. Is it because of units? The graph uses P/pKa, but what does that mean? Pressure per kiloPascal?
by Justin Vayakone 1C
Mon Jan 06, 2020 10:46 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Textbook question 5G.11
Replies: 2
Views: 22

Re: Textbook question 5G.11

5G.4 is all about Gibbs free energy which doesn't have anything to do with question 5G.11. Read through 5G.2 and you should be able to solve this problem. It's all partial pressures and mentions "Pure liquids and solids do not appear in K" (5G.2)
by Justin Vayakone 1C
Mon Jan 06, 2020 10:32 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Le Chatelier's and Endo/Exothermic
Replies: 5
Views: 38

Re: Le Chatelier's and Endo/Exothermic

Exothermic means energy is released while endothermic means energy is taken in, so exothermic:energy is a product, and endothermic: energy is a reactant. What I use to help me determine which is which is to think that endo sounds like "in", so it takes in energy. In exothermic reactions, e...
by Justin Vayakone 1C
Mon Jan 06, 2020 10:22 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Module: Equilibrium Part 3 Question 17
Replies: 3
Views: 31

Re: Module: Equilibrium Part 3 Question 17

If you can solve the problem, without the x^2 approximation, then do it that way. In this case, making x^2 equal to 0 is too much of an approximation. This is how I solved for x: (in file attachment)
Edit: Click on the image to see it right side up
by Justin Vayakone 1C
Mon Jan 06, 2020 9:44 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Very Large K
Replies: 12
Views: 64

Re: Very Large K

A very large K value means the reaction at equilibrium will heavily lean towards the products. The equilibrium constant is basically concentration of products over concentration of reactants, so for K to be a large value, the numerator (product) should be big while denominator (reactant) should be s...
by Justin Vayakone 1C
Mon Jan 06, 2020 9:33 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: HW Problem 5G.7
Replies: 4
Views: 41

Re: HW Problem 5G.7

My solutions manual uses partial pressures. Is your's seventh edition?
by Justin Vayakone 1C
Mon Jan 06, 2020 9:23 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 5l.7
Replies: 1
Views: 32

Re: 5l.7

Use this equation:
by Justin Vayakone 1C
Thu Dec 05, 2019 3:52 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: 6B 1
Replies: 2
Views: 33

Re: 6B 1

Doing log[.88] means the concentration was reduced by 12%, but the problem is asking for when it decreases to 12%. The equation for the initial pH is: pH = -log[HCl]_i Final pH: pH = -log(0.12[HCl]_i) To find the change, just do final pH - initial pH. Through some algebra, you should get -lo...
by Justin Vayakone 1C
Thu Dec 05, 2019 3:26 pm
Forum: Naming
Topic: naming
Replies: 5
Views: 43

Re: naming

Wait there was a worksheet he emailed us? When did he email it? I don't see it.
by Justin Vayakone 1C
Thu Dec 05, 2019 2:45 pm
Forum: Shape, Structure, Coordination Number, Ligands
Topic: water
Replies: 3
Views: 33

Re: water

When figuring out whether a ligand is monodenate or polydentate, we have to look at the lone pairs. Both of the lone pairs of water are on the oxygen right next to each other. When lone pairs are on the same atom, only one can bond because the other lone pair will point away from the bonding metal a...
by Justin Vayakone 1C
Thu Dec 05, 2019 2:12 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: CH4 versus CCl4 (Boiling Point)
Replies: 3
Views: 20

Re: CH4 versus CCl4 (Boiling Point)

When it comes to boiling point, we have to look at the intermolecular forces (e.g. dipole-dipole and london dispersion forces) not the intramolecular strength (like bond strength within a molecule). Since both CH4 and CCl4 are nonpolar, they both only have london dispersion forces for intermolecular...
by Justin Vayakone 1C
Thu Dec 05, 2019 2:03 pm
Forum: *Titrations & Titration Calculations
Topic: Wednesday Lecture
Replies: 3
Views: 42

Wednesday Lecture

In this lecture, what information do we have to know since Lavelle said some of the information was 14B material? Like for example, would we have to know stoichiometric point and the polyprotic acids and bases for the final?
by Justin Vayakone 1C
Thu Nov 28, 2019 2:36 pm
Forum: Hybridization
Topic: sp^2 Hybridization of Carbon
Replies: 2
Views: 20

Re: sp^2 Hybridization of Carbon

The answer is electron-electron repulsion. There is more repulsion when two electrons share an orbital making it less favorable and unstable. Therefore, to avoid this, one of the electrons in the s-orbital jumps to fill an empty p-orbital. Hope this helps! I'm still confused. In the first photo, th...
by Justin Vayakone 1C
Thu Nov 28, 2019 12:58 pm
Forum: Bronsted Acids & Bases
Topic: Acid vs Base
Replies: 4
Views: 47

Re: Acid vs Base

Yeah generally, molecules with OH will indicate a bronsted base and H for bronsted acids. For part a though, NH3 (ammonia) is a base. It will gain a proton to become NH4 (ammonium). Memorizing some of the common acids and bases will help in figuring out these molecules. B and D are the acids, while ...
by Justin Vayakone 1C
Thu Nov 28, 2019 12:47 pm
Forum: Hybridization
Topic: sp^2 Hybridization of Carbon
Replies: 2
Views: 20

sp^2 Hybridization of Carbon

Why is it that when we use sp^2 orbitals for Carbon, it involves three 2sp^2 and one 2p^1 orbital? Why not three 2sp^2 orbitals and one 2s^1 orbital?
by Justin Vayakone 1C
Thu Nov 28, 2019 12:28 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: Concentration
Replies: 2
Views: 28

Re: Concentration

It depends on what we are given. If a weak acid is involved, then we need to use the equilibrium constant (Ka) along with the given concentrations to make an ice table and calculate [H+] or [OH-], but I don't think we'll be facing these types of problems in this course. I haven't seen the practice p...
by Justin Vayakone 1C
Thu Nov 28, 2019 12:06 pm
Forum: Bronsted Acids & Bases
Topic: J17 Help
Replies: 2
Views: 41

Re: J17 Help

The problem is asking for equations involving a cation that is a weak acid reacting with water or an anion that is a weak base reacting with water, so they will look something like this: \textup{A}^++\textup{H}_2\textup{O}\rightleftharpoons \textup{A}+\textup{H}_3\textup{O}^+ or \textup{B}^-+\textup...
by Justin Vayakone 1C
Tue Nov 26, 2019 1:30 am
Forum: Hybridization
Topic: Unpaired electrons
Replies: 2
Views: 34

Re: Unpaired electrons

Yes each hybridized orbital represents bonds in the molecule. Let's take methane (CH4) for example. Because there are four regions of electron density around the Carbon atom, the s and p orbitals combine to create sp^3 orbitals: https://ibb.co/DGWMx8v Each of these sp^3 orbitals has an unpaired elec...
by Justin Vayakone 1C
Tue Nov 19, 2019 6:10 pm
Forum: Hybridization
Topic: Hybridization
Replies: 4
Views: 27

Re: Hybridization

Do all molecules use hybridized orbitals? No. Hybridized orbitals are only considered when necessary. For example, H2 molecules use the 1s orbital and don't need hybridized orbitals. You will have to look at the electron configuration of the atom to figure out if there are enough unpaired electrons...
by Justin Vayakone 1C
Mon Nov 18, 2019 11:07 pm
Forum: Hybridization
Topic: 2.57
Replies: 4
Views: 65

Re: 2.57

Image of Lewis Structure: https://ibb.co/FbB3TwM There are 2 electron densities around the right carbon, so sp hybridized orbitals are used for the triple bond. Here is the electron configuration shown by my epic MS paint skills: https://ibb.co/kQTh67k For the left carbon atom, sp^3 orbitals are use...
by Justin Vayakone 1C
Mon Nov 18, 2019 10:07 pm
Forum: Hybridization
Topic: Hybridization
Replies: 4
Views: 27

Re: Hybridization

Hybridization is just a model. It's used whenever necessary. If the current orbitals of a molecule don't have enough unpaired electrons to make all of the necessary bonds, then we use hybridized orbitals to explain for the lack of unpaired electrons. For example, a molecule like H2 won't have hybrid...
by Justin Vayakone 1C
Mon Nov 18, 2019 9:46 pm
Forum: Hybridization
Topic: Energy in Hybridization
Replies: 3
Views: 28

Re: Energy in Hybridization

Hybridization is just another addition to the bonding model to fix the issue that there weren't enough unpaired electrons to make all of the bonds, so I don't think energy of the molecule changes. But when it comes to the hybridized orbitals, I thought Lavelle said the energy of these orbitals were ...
by Justin Vayakone 1C
Mon Nov 18, 2019 9:23 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: 2E13 part d
Replies: 1
Views: 20

Re: 2E13 part d

https://imgur.com/a/AUAeA2Z Out of these three structures, the top right one is the best structure. Negative formal charge should be placed on the most electronegative atom if possible. In this case, O is more electronegative than N, so O should get the negative formal charge over N. Edit: Is the i...
by Justin Vayakone 1C
Mon Nov 18, 2019 9:05 pm
Forum: Dipole Moments
Topic: non polar dipole moments
Replies: 2
Views: 17

Re: non polar dipole moments

Nonpolar molecules can't have dipole-dipole interactions. Within each molecule, all of the dipoles cancel out, so there should be no dipole interaction between molecules. The only intermolecular force of nonpolar molecules is london dispersion forces.
by Justin Vayakone 1C
Thu Nov 14, 2019 4:39 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: why are double bonds equally weighted as single ones when drawing models?
Replies: 10
Views: 59

Re: why are double bonds equally weighted as single ones when drawing models?

I think double/triple bonds can distort the molecule geometry slightly, but the VSEPR model ignores that for simplicity I guess. It's just a model, so there are probably limitations and exceptions.
by Justin Vayakone 1C
Thu Nov 14, 2019 4:19 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: Vapor Pressure + IMF
Replies: 3
Views: 25

Re: Vapor Pressure + IMF

Imagine a closed container half filled with liquid. The temperature is then increased to the point where some of the liquid evaporated into gas. The gas particles colliding with the liquid creates what's called vapor pressure, pressure exerted by gas against it's solid or liquid phase. Vapor pressur...
by Justin Vayakone 1C
Thu Nov 14, 2019 3:59 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: 2F 9
Replies: 1
Views: 24

Re: 2F 9

Lavelle hasn't taught us hybridized orbitals yet, but it's pretty much combining orbitals into one type. For example, the two 2s electrons and two 2p electrons could combine to create 2sp^3 orbitals. Here is a table I made back from AP chem to help determine what hybridized orbitals are used. I woul...
by Justin Vayakone 1C
Thu Nov 14, 2019 2:10 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: Rotation of Polar Molecules in Dipole-Dipole Forces
Replies: 1
Views: 29

Re: Rotation of Polar Molecules in Dipole-Dipole Forces

Alright, I'll pretty much paraphrase the paragraph above the figure. In the gas phase, molecules are rotating rapidly. An assumption you can make is that there is no net interaction between the molecules because the attraction between opposite charges are canceled by the repulsion of like charges. T...
by Justin Vayakone 1C
Thu Nov 14, 2019 1:48 am
Forum: Ionic & Covalent Bonds
Topic: HW 3F #19
Replies: 1
Views: 30

Re: HW 3F #19

Vapor pressure is relative to boiling point. If the boiling point of one molecule is lower than a different molecule, then the first molecule will have more gas molecules, leading to more vapor pressure. To tie all of this in terms of IMFs, you can say a weaker IMF makes the molecules easier to brea...
by Justin Vayakone 1C
Wed Nov 06, 2019 4:39 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Mo and Ag exceptions
Replies: 4
Views: 79

Re: Mo and Ag exceptions

Yes, Mo and Ag want to have the half filled d-orbital over the full s-orbital, just like Cr and Cu above them.
by Justin Vayakone 1C
Wed Nov 06, 2019 4:35 pm
Forum: Photoelectric Effect
Topic: Photoelectric Effect vs. De Broglie
Replies: 8
Views: 109

Re: Photoelectric Effect vs. De Broglie

Do we use the De Broglie equation for the midterm review question 13a? Yes we do. We know this because we are using the wavelength of a potassium ion, which has mass. The speed of light equation (\textup{c}=\lambda \upsilon ) on the other hand would be used for situations involving electrom...
by Justin Vayakone 1C
Wed Nov 06, 2019 1:43 am
Forum: Properties of Light
Topic: Remembering order of EM radiation
Replies: 1
Views: 47

Re: Remembering order of EM radiation

Well for visible light, I use (ROY G BIV) to remember colors from weakest to strongest energy. Then to the left of visible light is infraRED light and to the right of visible light is ultraVIOLET light. Before infrared light and the start of the EM spectrum is radio waves followed by microwaves. I r...
by Justin Vayakone 1C
Wed Nov 06, 2019 12:28 am
Forum: Electron Configurations for Multi-Electron Atoms
Topic: 2A #21
Replies: 1
Views: 30

Re: 2A #21

The electron configuration of Ag is: [\textup{Kr}] 4\textup{d}^{10} 4\textup{s}^1 . Silver is one of the exceptions of the Aufbau principle because it is more stable to have a full d-subshell and one s-electron rather than a full s-subshell and 9 d-electrons. Since the question is asking for \textup...
by Justin Vayakone 1C
Tue Nov 05, 2019 11:46 pm
Forum: Photoelectric Effect
Topic: Photoelectric Effect vs. De Broglie
Replies: 8
Views: 109

Re: Photoelectric Effect vs. De Broglie

Do we use the De Broglie equation for the midterm review question 13a? Yes we do. We know this because we are using the wavelength of a potassium ion, which has mass. The speed of light equation (\textup{c}=\lambda \upsilon ) on the other hand would be used for situations involving electrom...
by Justin Vayakone 1C
Tue Nov 05, 2019 11:30 pm
Forum: Polarisability of Anions, The Polarizing Power of Cations
Topic: Polarizing Strength of Cations
Replies: 8
Views: 46

Re: Polarizing Strength of Cations

I think if the cation's radius is small, the electrons of the anion would be closer to the cation nucleus, creating a stronger pull and polarizing the electrons more easily.
by Justin Vayakone 1C
Tue Nov 05, 2019 11:16 pm
Forum: Trends in The Periodic Table
Topic: ionic radii
Replies: 4
Views: 36

Re: ionic radii

Actually, when we add electrons, the attractive pull between the nucleus and electrons are weaker. With more electrons to pull in, the force is weaker. There is also added repulsion between the electrons that cause them to be further apart, creating a larger radius.
by Justin Vayakone 1C
Tue Nov 05, 2019 11:03 pm
Forum: Trends in The Periodic Table
Topic: Half Filled Orbitals
Replies: 2
Views: 62

Half Filled Orbitals

So for number 9 in the Dino Nuggets Midterm Review, the UA's explained that oxygen actually has less ionization energy than nitrogen despite being on the right of nitrogen. I know it's because of half-filled orbitals being pretty stable, but does this concept apply to the periods below nitrogen and ...
by Justin Vayakone 1C
Fri Nov 01, 2019 10:28 pm
Forum: Lewis Structures
Topic: Expanded Shells
Replies: 3
Views: 54

Re: Expanded Shells

Well d-subshells can hold 10 electrons, so I guess 18 could be the max? I don't think we need to worry about a max because we won't ever get to that point.
by Justin Vayakone 1C
Wed Oct 30, 2019 6:07 pm
Forum: Ionic & Covalent Bonds
Topic: Homework 2A.11
Replies: 2
Views: 26

Re: Homework 2A.11

If we were looking at \textup{M}^{2+} , then your answer of Iron and Manganese would be correct, but we are looking at metal cations with a 3+ charge or in other words elements that lose 3 electrons. Let's take a look at Iron. The electron configuration of Iron is [\textup{Ar}]3\textup{d}^64\textup{...
by Justin Vayakone 1C
Wed Oct 30, 2019 5:33 pm
Forum: Electronegativity
Topic: 3F. 5C
Replies: 1
Views: 33

Re: 3F. 5C

The size of the atoms largely contribute to the melting point of molecules. Fluorine may have stronger dipole-dipole attraction than Iodine, but that difference in electronegativity is trumped by the difference in size. Iodine is bigger than Fluorine, meaning more electrons and stronger London Dispe...
by Justin Vayakone 1C
Wed Oct 30, 2019 5:20 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: 1E.25 Part C
Replies: 2
Views: 19

Re: 1E.25 Part C

In my solution manual, the answer to part c is ns^2 (n-1)d^3. That's the answer you said in your post right? ns^2 is the 2 valence electrons, and the (n-1)d^3 is the outermost d-orbital electrons.
by Justin Vayakone 1C
Wed Oct 30, 2019 5:10 pm
Forum: Ionic & Covalent Bonds
Topic: ground state electron configuration
Replies: 1
Views: 21

Re: ground state electron configuration

For part a and b, keep in mind that the 4s orbital is higher in energy than the 3d orbital. The question is asking for metal cations with a 2+ charge, meaning the metal element will lose 2 electrons. Part a and b involve electron configurations with electrons in the 3d and 4s orbitals. Since 4s is h...
by Justin Vayakone 1C
Wed Oct 30, 2019 4:51 pm
Forum: Molarity, Solutions, Dilutions
Topic: Practice Question for Midterm
Replies: 2
Views: 80

Re: Practice Question for Midterm

To solve this problem, we can use the \textup{M}_1{}\textup{V}_1 = \textup{M}_2{}\textup{V}_2 equation. To find the original molarity, we have to first convert the 5.00 g of KMnO4 into moles using KMnO4 molar mass ( 5.00\textup{g} \textup{KMnO}_4\div158.04\textup{g/mol}= 0.0316 \textup{mol KMnO}_4 )...
by Justin Vayakone 1C
Mon Oct 21, 2019 10:29 pm
Forum: Lewis Structures
Topic: Homework 2B.3 d
Replies: 2
Views: 36

Re: Homework 2B.3 d

BrF3 has 28 electrons. This molecule is one of the exceptions to the octet rule. Br can have an expanded octet. Br will be the central atom with single bonds to each F and 2 lone pairs to itself.
by Justin Vayakone 1C
Mon Oct 21, 2019 10:20 pm
Forum: Trends in The Periodic Table
Topic: Ionization Energies Trend
Replies: 5
Views: 31

Re: Ionization Energies Trend

I guess the increase in nuclear charge trumps the electron shielding. Maybe because when increasing across a period, the electrons added are valence electrons, so I think the electron shielding really isn't increasing much.
by Justin Vayakone 1C
Mon Oct 21, 2019 10:06 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Ground state electron configuration of ions
Replies: 4
Views: 33

Re: Ground state electron configuration of ions

No, the ground state of an ion is the ion itself. For example, the ground state of Ca+ is Ca+.
by Justin Vayakone 1C
Mon Oct 21, 2019 5:58 pm
Forum: Properties of Light
Topic: 1B.15 --> c=λv vs. λ= h/mv
Replies: 2
Views: 31

Re: 1B.15 --> c=λv vs. λ= h/mv

The equation is \textup{c}=\lambda\cdot v . The v actually represents frequency, not velocity. The units of the equation are (\textup{m/s}) = (\textup{m}))(s^{-1}) . As you mentioned, the units don't work if you plug in velocity into that equation. Checking units of each ...
by Justin Vayakone 1C
Mon Oct 21, 2019 5:45 pm
Forum: Einstein Equation
Topic: units
Replies: 1
Views: 40

Re: units

It stands for kilo-electron volt or 1000 electron volts. The SI unit of this is joules. The conversion is
by Justin Vayakone 1C
Sun Oct 20, 2019 8:42 pm
Forum: Quantum Numbers and The H-Atom
Topic: 1D.21
Replies: 5
Views: 35

Re: 1D.21

The question is asking for a specific subshell. When we are given an n and l value, we will have a subshell. For part a) since n=5 and l=2, we know the subshell is 5d. Subshell notation is simply just the n value followed by s, p, d, or f subshells. l=0 is the s-subshell, l=1 is the p-subshell, l=2 ...
by Justin Vayakone 1C
Sun Oct 20, 2019 8:26 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: 1E-11
Replies: 4
Views: 39

Re: 1E-11

Electron configuration is just an easy way to see where all of the electrons are in the orbitals. When we are stating all of the orbitals, we start from lowest energy to highest energy. The periodic table is a great way to figure out what orbitals are occupied. Here is a link to see the periodic tab...
by Justin Vayakone 1C
Sun Oct 20, 2019 8:00 pm
Forum: Quantum Numbers and The H-Atom
Topic: 1E. 1
Replies: 4
Views: 45

Re: 1E. 1

Actually, all four of the values would still increase if the electron was in a hydrogen atom. All of the reasoning for the Lithium electron is the same for a hydrogen atom. In both cases, the the electron moves from an s-orbital to a p-orbital, which means the energy and n value is increased. When n...
by Justin Vayakone 1C
Sun Oct 20, 2019 7:30 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: 4s orbitals and 3d orbitals
Replies: 2
Views: 34

Re: 4s orbitals and 3d orbitals

When elements have z\leq 20 , then the 4s orbital is less in energy than 3d meaning 4s is written first. In other words, elements up to Calcium will have the 4s state come before 3d. But for elements after calcium in the periodic table, the 4s state is actually higher in energy than 3d. For example ...
by Justin Vayakone 1C
Sun Oct 20, 2019 7:15 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: Problem 1a.11
Replies: 3
Views: 35

Re: Problem 1a.11

Each line represents energy released by the movement of a hydrogen electron from one energy level to a lower energy level. Each energy level is represented by a principal quantum number (n). What's common among each series is that their lines all involve the electron being in the same final energy l...
by Justin Vayakone 1C
Thu Oct 10, 2019 4:39 pm
Forum: Student Social/Study Group
Topic: Pen or Pencil for Homework Problems?
Replies: 8
Views: 72

Pen or Pencil for Homework Problems?

Does it matter if these practice problems are in pen or pencil?
by Justin Vayakone 1C
Tue Oct 08, 2019 11:20 pm
Forum: Trends in The Periodic Table
Topic: Chemistry Community
Replies: 8
Views: 79

Re: Chemistry Community

I think they are due every Sunday.
by Justin Vayakone 1C
Tue Oct 08, 2019 7:28 pm
Forum: Administrative Questions and Class Announcements
Topic: Finding Textbook Questions on the Quantum World
Replies: 4
Views: 46

Re: Finding Textbook Questions on the Quantum World

In the seventh edition of the textbook, the 1A problems are right after Topic 1A on pg 9. Regarding turning in homework questions, I believe you can turn in questions just from Review of Chemical Principles if you want, or only from Quantum world, or a mix of the two.
by Justin Vayakone 1C
Tue Oct 08, 2019 7:02 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Spectral Lines
Replies: 3
Views: 61

Re: Spectral Lines

Location on the electromagnetic spectrum doesn't exactly determine which series a line belongs to. Each line represents energy released by the movement of a hydrogen electron from one energy level to a lower energy level. Each energy level is represented by what's called the principal quantum number...
by Justin Vayakone 1C
Tue Oct 08, 2019 6:14 pm
Forum: Accuracy, Precision, Mole, Other Definitions
Topic: Fundamentals E17
Replies: 2
Views: 69

Re: Fundamentals E17

The question is asking for the sample with more moles. In part a, you would convert both the 75g indium and 80g tellurium into their respective moles by dividing the substance molar mass. 75g \textrm{In} \div 114.82g/mol = 0.65 mol \textrm{In} 80g \textrm{Te} \div 127.60g/mol = 0.63 mol \textrm{Te} ...
by Justin Vayakone 1C
Tue Oct 08, 2019 5:42 pm
Forum: Photoelectric Effect
Topic: Topic 1B Quantum Theory Example
Replies: 2
Views: 22

Re: Topic 1B Quantum Theory Example

The mass of an electron is 9.109 x 10^(-31) kg. If we needed to use this constant on a test, I'm sure we would be given the value on a sheet that has all of the formulas and constants.
by Justin Vayakone 1C
Tue Oct 08, 2019 5:23 pm
Forum: Properties of Electrons
Topic: Electron Gaining/Losing energy
Replies: 5
Views: 281

Re: Electron Gaining/Losing energy

Just to add to the topic, when an electron goes down to lower energy levels, it releases energy in the form of photons.
by Justin Vayakone 1C
Thu Oct 03, 2019 6:02 pm
Forum: Limiting Reactant Calculations
Topic: Alternate way of solving for limiting reactant
Replies: 2
Views: 42

Re: Alternate way of solving for limiting reactant

It doesn't matter which product you use. As long as you convert each reactant into the same product, you will be able to compare how much product each reactant produced. Whichever reactant produced less of the product is the limiting reactant.
by Justin Vayakone 1C
Tue Oct 01, 2019 12:38 am
Forum: Limiting Reactant Calculations
Topic: Limiting Reactant Calculation post-module question
Replies: 4
Views: 45

Re: Limiting Reactant Calculation post-module question

Since C6H9Cl3 is the limiting reactant and AgCl is the product we want to calculate, we need to focus on the molar ratio between them, which is 1 to 1. Because the molar ratio is 1 to 1 (meaning they have the same coefficients in the balanced equation), 0.004 mol C6H9Cl3 produces 0.004 mol AgCl. The...
by Justin Vayakone 1C
Tue Oct 01, 2019 12:22 am
Forum: Limiting Reactant Calculations
Topic: How to Find a Mass of A Product
Replies: 4
Views: 39

Re: How to Find a Mass of A Product

We can't find the mass of a product with only molar mass. We need to be given some more information, like moles of reactants or products for example. If moles of a product is given, we multiply that number by the molar mass of that product to find the mass in grams. If moles of reactants are given, ...
by Justin Vayakone 1C
Mon Sep 30, 2019 11:57 pm
Forum: Accuracy, Precision, Mole, Other Definitions
Topic: Accuracy vs Precision
Replies: 10
Views: 125

Re: Accuracy vs Precision

A simple way for me to figure out the difference between these two terms is to think of precision as consistency and accuracy as the bullseye of a target.
by Justin Vayakone 1C
Mon Sep 30, 2019 11:45 pm
Forum: Molarity, Solutions, Dilutions
Topic: Help on G25
Replies: 3
Views: 65

Re: Help on G25

I see this as a dilution problem where we can use MV = MV. To represent the volume doubling 90 times, we can use this expression: (0.10)(2^90)L. The equation should be (0.10M)(0.01L) = (? M) [(0.10)(2^90)L] . By isolating the final molarity, we can find its value. Then multiply that value by 0.01L t...
by Justin Vayakone 1C
Mon Sep 30, 2019 11:07 pm
Forum: Administrative Questions and Class Announcements
Topic: Advice from a Medical Student - Part II [ENDORSED]
Replies: 130
Views: 3410

Re: Advice from a Medical Student - Part II [ENDORSED]

Thank you for sharing your path to becoming a doctor. I feel overwhelmed with how long and difficult the journey through medicine is, but these types of stories help give me hope and inspiration.

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