## Search found 102 matches

Sat Mar 14, 2020 1:11 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Equilibrium and Kinetics
Replies: 2
Views: 65

### Re: Equilibrium and Kinetics

One way they are connected is through the equilibrium constant K, which can be written in terms of the rate constants K = kf/kr. This means that if the forward rate reaction is larger, then so is the equilibrium constant and the final concentrations of the products.
Sat Mar 14, 2020 12:52 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Delta S
Replies: 8
Views: 249

### Re: Delta S

Delta S total is always 0 or greater, as it accounts for the entropy of everything and entropy always increases. It is made up of Ssys and Ssur, depending on what you define as your system. The 2 added together equal S total.
Sat Mar 14, 2020 12:24 pm
Forum: First Order Reactions
Topic: First order rxns
Replies: 6
Views: 161

### Re: First order rxns

There are a couple ways to determine this. You could graph the log of the concentration over time is linear. Or, you can check the rate to check if it is dependent only on the concentration of the reactant to the 1st power. For example, if you double the initial concentration, then the initial rate ...
Sat Mar 14, 2020 12:22 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Midterm 6A
Replies: 1
Views: 110

### Re: Midterm 6A

The stronger the bond, the more energy is needed to break them. As the reaction is exothermic, you will need to input energy to get it back to the initial state so the final products bonds are stronger and more stable.
Wed Mar 11, 2020 11:01 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: 7D.7
Replies: 1
Views: 55

### Re: 7D.7

You know that the equilibrium constant K can be written as kforward/kreverse. So you can just divide the 2 numbers to find the constant.
Mon Mar 02, 2020 10:58 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 6L.7A
Replies: 1
Views: 42

### Re: 6L.7A

In the back of the book, you can find a half reaction that looks like AgBr + e- > Ag + Br-. You can use that as the half reaction for Br-. Then for Ag, you can use the usual Ag > Ag+ + e-.
Mon Mar 02, 2020 10:54 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: HW 6.57
Replies: 6
Views: 188

### Re: HW 6.57

By definition, the Ka for the reaction would be in the form [H+][F-]/[HF]. When you end up adding the F- half reaction and calculate K, the equation for K will look like [H+]^2, as in the half reaction below, there is a coefficient of 2 in front of H+. To solve for Ka, you can just square root.
Mon Mar 02, 2020 10:41 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 6M13 part d
Replies: 1
Views: 29

### Re: 6M13 part d

The reduction potential you looked up for NO3- was to turn it into NO. For the reduction potential for NO3- to NO2, E= 0.80
Mon Mar 02, 2020 10:38 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: Water with dissolved ions
Replies: 1
Views: 56

### Re: Water with dissolved ions

This is because the ions in the water have charges, and when you pass a current through it, they can separate and move around when pure water cannot. These ions allow charge to flow which is what conducting means.
Mon Mar 02, 2020 10:30 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Relationship between E° + K
Replies: 1
Views: 36

### Re: Relationship between E° + K

If Enot is greater than 0, from the equation G= -nFE = -RTlnK, so Enot = RT/nF lnK. If E not is positive, then so does lnK, which only occurs when K>1.
Wed Feb 26, 2020 9:57 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: 6N3.B
Replies: 2
Views: 64

### Re: 6N3.B

You are right that zinc is the reactant, but that means that Zn2+ is the product, so that goes on top when calculating Q
Wed Feb 26, 2020 1:02 am
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: 6.57
Replies: 3
Views: 113

### Re: 6.57

You can relate E to Gnot using Gnot = -nFE. From there, relate Gnot to K using Gnot = -RT lnK. In this case, the reaction given makes K equal to Ka
Wed Feb 26, 2020 12:58 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 6L.1
Replies: 2
Views: 55

### Re: 6L.1

You have to write down the half reactions to see how many electrons are transferred, and use that for your n.

In this case 2 Ce 4+ + 2e- => 2 Ce3+, so n = 2 moles of electrons
Tue Feb 25, 2020 8:34 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 6M.11D
Replies: 1
Views: 56

### Re: 6M.11D

Dr. Lavelle writes it like that to show the flow of electrons and to make it more clear for us, but it isn't convention and you can write it in either order.
Tue Feb 25, 2020 8:31 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 6L9
Replies: 2
Views: 78

### Re: 6L9

You have to look at the oxidation numbers and the half reactions will involve the ones where it is changing. So, Cl stays at - and doesn't change, so it doesn't need a half reaction. On the other hand, iron is going from 2+ to 3+, and permanganate turns into a manganese ion which also changes the ox...
Sun Feb 23, 2020 4:20 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: movement e
Replies: 3
Views: 33

### Re: movement e

There is a charge difference between the 2 cells. Because there is a high concentration of electrons in the anode, it wants to reach equilibrium and move towards the low concentration of electrons in the cathode. It does this through the wire connecting the 2 cells, as the metals allows for the flow...
Thu Feb 20, 2020 1:04 pm
Forum: Balancing Redox Reactions
Topic: Oxidation Numbers
Replies: 5
Views: 80

### Re: Oxidation Numbers

Typically, you assume that hydrogen has an oxidation state of 1+, and oxygen 2-. From there, you can calculate the unknown oxidation numbers based on the total net charge of the molecule. For ionic compounds, you can use their charges.
Wed Feb 19, 2020 1:15 am
Forum: Balancing Redox Reactions
Topic: 6K.3 Part D
Replies: 2
Views: 69

### Re: 6K.3 Part D

I think its a typo where the Cl2 on the right side should be a Cl-, at least from what I found from previous years.
Wed Feb 19, 2020 1:12 am
Forum: Balancing Redox Reactions
Topic: Balancing the reduction of o3->o2 in basic soln
Replies: 1
Views: 35

### Re: Balancing the reduction of o3->o2 in basic soln

Before the last step, there are 2 Hydrogen atoms present in the H2O. This means you must add 2 OH- and H2O on opposite sides to balance. Also, you can cancel out the waters on both sides until there is only water on 1 side remaining.
Mon Feb 17, 2020 3:06 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: Constant R
Replies: 15
Views: 390

### Re: Constant R

Usually when you are solving for pressure or volume using PV=nrt you use the one with liters and atm (.08205). If you are solving for work, it is more common to use the one with Joules in the units (8.3145).
Tue Feb 11, 2020 7:28 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 4J.13 stability
Replies: 1
Views: 35

### Re: 4J.13 stability

Yes, if you write out the equation to form the molecules, then you can compare the Gibbs free energy for all of them. For example, H2 + 2C + N2 = 2HCN. The reaction with the lowest Gibbs free energy means that it will be the most stable, as the product will form spontneously.
Tue Feb 11, 2020 1:29 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Test #1 Problem #2
Replies: 2
Views: 129

### Re: Test #1 Problem #2

You can find the equilibrium quotient given the initial conditions and compare it with equilibrium. In this case, Q = [H2][CO2]/[CO][H2O]. You can input the concentrations into it and see if it's larger or smaller than K =31.4. If larger, then there is too much product and the reaction shifts to the...
Tue Feb 11, 2020 1:21 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Isothermal irreversible reactions
Replies: 3
Views: 85

### Re: Isothermal irreversible reactions

Because internal energy is a state function, that means that it will be the same regardless of the path taken. If the initial and final temperature are the same, then U = 0.
Tue Feb 11, 2020 1:16 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Calculating Change in Entropy for Phase Changes
Replies: 1
Views: 38

### Re: Calculating Change in Entropy for Phase Changes

This is because you have to change the phase of the substance. The 2 different phases will also have different molar entropies.
Mon Feb 10, 2020 4:13 pm
Forum: Calculating Work of Expansion
Topic: Reversible vs irreversible expansions
Replies: 2
Views: 83

### Re: Reversible vs irreversible expansions

In reversible expansions, the outside pressure is always equal the the internal pressure. In irreversible expansions, there is a constant external pressure. This means that the amount of work done by the 2 systems will be different, with the reversible expansion doing the maximum amount of work.
Mon Feb 10, 2020 4:11 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: calculating concentrations
Replies: 1
Views: 25

### Re: calculating concentrations

You can calculate the value of x disregarding when you subtract it. Then, you compare the value of x with what you would subtract it from. One example is how much of a substance dissociates. If your equation looks like x/(.4-x)=Ka, calculate x and see if it is less than 5% of .4.
Tue Feb 04, 2020 12:03 am
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Problem 4D.3
Replies: 2
Views: 54

### Re: Problem 4D.3

Since you are using a bomb calorimeter, volume is not changing so dU = q + w, where w = PdV. Since volume doesn't change, you can cancel out the w term and change in internal energy is just q. You still have to subtract the heat capacity times the change in pressure to account for the calorimeter.
Mon Feb 03, 2020 11:57 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: 4F.7 - no given Cp constant?
Replies: 3
Views: 61

### Re: 4F.7 - no given Cp constant?

In an earlier section, they mention that you can find the Cp and Cv of an ideal gas through derivations. The values are on page 265 and 266 and should be 3/2 R and 5/2 R.
Mon Feb 03, 2020 11:35 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Ch 5I question #19
Replies: 2
Views: 56

### Re: Ch 5I question #19

Yes, since for H2 to be used, HI must form (twice as much). No matter is created or destroyed, so it must have come from the reactants given.
Mon Feb 03, 2020 11:32 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: microstates
Replies: 1
Views: 24

### Re: microstates

If you have 3 particles in a 2 state system, then each particle can have 1 of 2 states. To find the total number of microstates, you multiply the number of possible states the number of particles times. In this case, it would be 2^3. For 1 mol of an atom, it would be 2^Avagadros number. You can use ...
Mon Feb 03, 2020 11:28 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: 4A.11
Replies: 5
Views: 90

### Re: 4A.11

In this case, heat capacity is only defined by energy put in vs change in temperature, q/T. This is different from specific heat, which includes units for mass. The amount of substance matters.
Tue Jan 28, 2020 12:41 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Enthalpy at a constant pressure vs at a constant volume
Replies: 1
Views: 51

### Re: Enthalpy at a constant pressure vs at a constant volume

For constant volume, you are calculating delta U instead of H, which is the change in internal energy, and to convert between the 2, you use the equation ΔH=qv+ΔngRT. For constant pressure, you can just use the equation ΔH=qv= ΔT*Csp*m
Tue Jan 28, 2020 9:13 am
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Heat capacity units
Replies: 1
Views: 15

### Re: Heat capacity units

No, as a change in 1 degree Kelvin is equal to 1 degree Celsius.
Tue Jan 28, 2020 12:09 am
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Constant Pressure Calorimeter
Replies: 1
Views: 26

### Re: Constant Pressure Calorimeter

This is because for constant volume calorimeters, you are measuring delta U instead. Constant pressure reactions results in a change in volume, so work is done which needs energy, which delta H does not include. To get a better answer for change in energy, volume needs to be fixed to get delta U ins...
Tue Jan 28, 2020 12:04 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 4D1
Replies: 2
Views: 51

### Re: 4D1

You need to divide by 4 since there are 4 moles in the product side, which means that when you form 4 moles of the product, 358.8 kJ is released. That means when 1 mole is formed, only 358.8/4 kJ of energy is released. You want the coefficient of the product you are interested in to be 1. Delta H is...
Tue Jan 28, 2020 12:01 am
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: homework question 4A.3
Replies: 6
Views: 94

### Re: homework question 4A.3

Work is equal to negative pressure times change in volume. You can use the equation to calculate all the numbers.
Sat Jan 25, 2020 4:58 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: standard enthalpy
Replies: 3
Views: 44

### Re: standard enthalpy

The definition of standard enthalpy is the change in enthalpy to form the compound from the most stable form of its elementary parts. So if you were trying to form O2, then the reaction would just be O2 -> O2, and since they are the same thing, it takes no energy.
Thu Jan 23, 2020 1:27 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Ka and Kb
Replies: 10
Views: 113

### Re: Ka and Kb

The 2 constants would have the same inequality. If either Kb or Ka is weak, then the value should be less than 10^-3. Looking at the equation,

Ka = [H+][A-]/[HA], so the smaller Ka, is, the less it dissociates and the lower the concentration of H+
Wed Jan 22, 2020 11:16 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: pressure and volume
Replies: 3
Views: 50

### Re: pressure and volume

Pressure is inversely proportional to volume as seen by the equation PV = nRT. Given that moles and temperature stay constant, when pressure increases, volume must decrease. It's more commonly thought of that you halve the volume to double the pressure rather than changing the pressure.
Wed Jan 22, 2020 11:12 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Determining pH
Replies: 3
Views: 35

### Re: Determining pH

Weak acids and bases are just those not considered strong. I think you just need to memorize them. The strong acids include most halogens and some oxoacids like HCl, HI, HBr, HClO4, HNO3, and H2SO4. The strong bases include Group 1 and heavy group 2 hydroxides, like LiOH and B(OH)2.
Mon Jan 20, 2020 11:38 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 6D.15
Replies: 2
Views: 43

### Re: 6D.15

You have to look at the ions created: Al3+ and Cl-. Cl- is the conjugate for a strong acid so it does not contribute to the pH. But, Al3- can form Al(OH)3, and some will form. You have to write the Kb equation to see how the concentration of OH- changes to find the final pOH and therefore the pH as ...
Mon Jan 13, 2020 3:10 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: 5J.3
Replies: 5
Views: 53

### Re: 5J.3

When you remove NO, the system wants to balance out agian and produce more. This means that the forward reaction proceeds and NH3 is used up.
Mon Jan 13, 2020 3:05 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Autoprotolysis
Replies: 6
Views: 88

### Re: Autoprotolysis

It can occur in other molecules as well, but I dont think it's relevant to the class. For example, ammonia, NH3, can self dissociate into NH4+ and NH2-. The resulting reaction would have a K value of 10^-30.
Mon Jan 13, 2020 12:35 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Water in Equilibrium Constant
Replies: 2
Views: 24

### Re: Water in Equilibrium Constant

When the solutions are aqueous or water is a liquid, it is left out of the KC equation. Only if it is not in excess or it is in a gaseous phase do you add it.
Mon Jan 13, 2020 12:13 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Kw calcuation
Replies: 5
Views: 56

### Re: Kw calcuation

Water is the solvent for the equation, and there is a huge excess of it compared to the OH- and H+ ions. This means that even if we did include it, the concentration barely changes.
Mon Jan 13, 2020 12:11 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 5H.3
Replies: 2
Views: 37

### Re: 5H.3

The equation given is actually the combination of 2 equations, one for HCl and one for BrCl. If you write out the KC equilibrium equation, you can add a Cl to the top and bottom without changing the value. Then, you can group the terms in one equation together and the final answer should be the prod...
Wed Jan 08, 2020 12:52 am
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: 5J.11b Halogens
Replies: 1
Views: 58

### Re: 5J.11b Halogens

Halogens themselves don't tell us if the reaction is exo or endothermic. However, breaking bonds requires energy so going from X2 to 2X is endothermic. This means that increasing the temperature will shift the equilibrium towards the products. The fact it is a halogen doesn't really matter.
Wed Jan 08, 2020 12:50 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 5H.1
Replies: 1
Views: 28

### Re: 5H.1

When you write out the K value, you take the concentrations of the products and reactants raise it to the coefficients power. If you write out the actual value for c for example, K = [NH3]^4/[N2]^2[H2]^6 compared to the original [NH3]^2/[N2][H2]^3. If you square the original K value, you get the new...
Tue Jan 07, 2020 9:20 am
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Pressure changes to equilibrium equations
Replies: 5
Views: 99

### Re: Pressure changes to equilibrium equations

Le Chatliers principle states that a system wants to reduce the stress that is applied. As the volume was decreased so the partial pressures were increased, the equilibrium wants to shift to the side with less moles of gas to reduce the pressure. This means it forms more product and favors the right.
Tue Jan 07, 2020 9:16 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Composition of equilibrium mixture
Replies: 2
Views: 43

### Re: Composition of equilibrium mixture

Usually when you are dealing with a cubic, the Kc constant is incredibly small (<10^-4) and we can assume the dissociation is negligible. Then, when writing the Kc, you can replace any term such as .4-x or .02-2x as just .4 or .02. This simplifies the equation into usually a much easier equation to ...
Mon Jan 06, 2020 8:11 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Equilibrium constant v. Reaction quotient
Replies: 3
Views: 64

### Re: Equilibrium constant v. Reaction quotient

Equilibrium constant is the value only when the reaction is at equilibrium and will be constant given fixed conditions. Reaction quotient is the value at any time and the value varies depending on the concentrations.
Mon Dec 02, 2019 10:57 pm
Forum: Properties & Structures of Inorganic & Organic Bases
Topic: Oxides of main group elements
Replies: 2
Views: 121

### Re: Oxides of main group elements

Yes, both are considered bases. Most metal oxides and hydroxides are considered bases.
Mon Dec 02, 2019 10:53 pm
Forum: Naming
Topic: Latin names of Elements
Replies: 1
Views: 32

### Re: Latin names of Elements

I think those are referring to complex ions with negative charges. If it does, then the ending changes to an ate, like cobalt to cobaltate. The latin names refers only to certain elements, like gold, silver, tin, iron, and copper.
Mon Dec 02, 2019 10:28 pm
Forum: Bronsted Acids & Bases
Topic: general q
Replies: 2
Views: 73

### Re: general q

pH is determined by the concentration of H+ ions in a solution. This means that anything that produces H+ or OH- ions can have pH, whether it be ions or salts or even gases like HCl.
Mon Dec 02, 2019 10:24 pm
Forum: Identifying Acidic & Basic Salts
Topic: HW 6D11
Replies: 7
Views: 144

### Re: HW 6D11

The H2O comes from the fact that the solution is aqueous so the salt is dissolved in water and interacts with the water molecules. The H2O donates a proton, turning it into a hydroxide (OH-) ion.
Mon Dec 02, 2019 10:17 pm
Forum: Lewis Acids & Bases
Topic: Acids and Bases 6A.13
Replies: 2
Views: 64

### Re: Acids and Bases 6A.13

Lewis bases donate electrons in a reaction with a lewis acid. In this case, F- and H- all have full set of electrons, which can form bonds with positive cations. In this sense, they are "donating" an electron pair. BF3 is missing an electron pair for a full octet, so it will form a coordin...
Tue Nov 26, 2019 9:27 pm
Forum: Naming
Topic: Bis,tris, etc
Replies: 6
Views: 127

### Re: Bis,tris, etc

You use bis and tris when the ligand already has a bi or prefix in front of it. For example, you would say bisethylenediammine as en already has a "di" in it. You also use bis and tris when the ligand is polydentate like EDTA and oxalate.
Tue Nov 26, 2019 9:23 pm
Forum: Shape, Structure, Coordination Number, Ligands
Topic: 17.37 Determine coordination number
Replies: 4
Views: 1848

### Re: 17.37 Determine coordination number

EDTA is ethylenediaminetetraacetate, which is an ethylenediamine (en) with 4 acetate groups attached to it. You kind of just have to memorize that it is hexadentate.
Tue Nov 26, 2019 9:20 pm
Forum: Shape, Structure, Coordination Number, Ligands
Topic: 9C.5
Replies: 1
Views: 44

### Re: 9C.5

If a ligand is polydentate, that means that there are 2 or more atoms on the molecule with a lone pair that can bind to the metal cation. For example, in a), there are 3 nitrogen atoms with lone pairs that could bind with a cation. However, in the case of water, even though oxygen has 2 lone pairs, ...
Mon Nov 25, 2019 10:51 pm
Forum: Shape, Structure, Coordination Number, Ligands
Topic: coordination number
Replies: 3
Views: 65

### Re: coordination number

en contributes 2 binding sites, so 2 of them create 4 bonds. That makes 6 total bonds if you include the 2 bonds from Calcium meaning it has a coordination number of 6.
Mon Nov 25, 2019 10:49 pm
Forum: Naming
Topic: Order of Ligands
Replies: 4
Views: 111

### Order of Ligands

Does it matter what order the ligands come in for the actual formula. For example, are [Co(NH3)5Cl]+ and [CoCl(NH3)5]+ both acceptable?
Tue Nov 19, 2019 1:10 am
Forum: Determining Molecular Shape (VSEPR)
Topic: 2E.1
Replies: 4
Views: 84

### Re: 2E.1

One example where the molecule is linear is in AX2E3, as the electron repulsions cancel out. All of the equatorial domains are electron pairs and the 2 axial ones are in a straight line.
Mon Nov 18, 2019 11:27 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: 3.F.1 c and d
Replies: 1
Views: 39

### Re: 3.F.1 c and d

If you look at the lewis structures for the molecules, you will see that there is a lone pair of electrons on the sulfur on SO2, which creates a dipole moment as the electron density is higher in a lone pair than a bonding region. In addition, HSeO4 has an OH group which creates hydrogen bonding as ...
Mon Nov 18, 2019 1:32 pm
Forum: Sigma & Pi Bonds
Topic: Sigma and Pi Bonds
Replies: 4
Views: 129

### Re: Sigma and Pi Bonds

In the majority of cases, this holds true, especially for what we are doing in this class. The first bond is considered a sigma bond and everything after is pi. In transition metals, overlapping d orbitals can sometimes lead to delta bonds, but we don't need to know about them.
Mon Nov 18, 2019 1:26 pm
Forum: Hybridization
Topic: 2F.7
Replies: 2
Views: 33

### Re: 2F.7

AsF3 also has a lone electron pair, which counts as an electron domain. This makes for 4 total domains and therefore sp3 hybridization.
Mon Nov 18, 2019 1:25 pm
Forum: Dipole Moments
Topic: Dipole moments
Replies: 4
Views: 95

### Re: Dipole moments

I think Dr. Lavelle mentioned we would be using convention, which is drawing it towards the negative charge.
Mon Nov 11, 2019 10:44 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: when there are three atoms
Replies: 1
Views: 34

### Re: when there are three atoms

In general, the atom and lone pairs shouldn't change the overall shape too much, but the angles between them might change slightly, like how lone pairs repel more than bonding domains. Also, if you are talking about IO2F2-, then I think the VESPR formula would give you a seesaw shape as there is als...
Mon Nov 11, 2019 10:38 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Interactions between Ions and Molecules
Replies: 1
Views: 25

### Re: Interactions between Ions and Molecules

Ion Ion and Ion dipole molecules will have a positive and negative part (or partially positive and negative). This attraction between the positive and negative charges defines the interaction. Ion Ion are only found between ionic compounds as they have clear whole charges while ion dipole is the att...
Mon Nov 11, 2019 9:58 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: VSEPR formula for molecules of the same atom
Replies: 3
Views: 38

### Re: VSEPR formula for molecules of the same atom

The VESPR for I3- would be treated the same way as other molecules.
It would be in the form AX2E3, which will make it a linear molecule as the lone pairs would go in the equatorial domains.
Mon Nov 11, 2019 1:35 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: intermolecular forces/polarity
Replies: 2
Views: 43

### Re: intermolecular forces/polarity

You have to draw out the Lewis structure to see if there is a dipole moment. In AsF3, the molecule is symmetric with no lone pairs so even if there is a polar bond between As and F, they are pulling electrons all in different directions so the net molecule has no polarity. On the other hand, AsF5 ha...
Mon Nov 11, 2019 1:32 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: types of intermolecular forces
Replies: 5
Views: 47

### Re: types of intermolecular forces

If the molecule is polar, then it will have dipole dipole forces. You can check this using the electronegativity difference and the molecular shape. For example, H and Cl have a very large difference in electronegativity so HCl would be a polar bond and therefore form dipole dipole interactions.
Mon Nov 04, 2019 11:42 pm
Forum: SI Units, Unit Conversions
Topic: Formula Units
Replies: 2
Views: 145

### Re: Formula Units

Formula Units are the equivalent of molecules or atoms for ionic compounds. Its just a term to refer to 1 unit of an ionic compound and you would use Avagadro's number to convert from moles to formula units.
Mon Nov 04, 2019 11:38 pm
Forum: Quantum Numbers and The H-Atom
Topic: How many orbitals?
Replies: 2
Views: 83

### Re: How many orbitals?

For b and d, they provide the ml quantum number, which corresponds to an orbital. This means that there is only 1. C asks to count how many orbitals are possible for n=2. You can find this by counting how many different l and ml combinations exist for n=2. You know l can be either 0 or 1 and if it i...
Mon Nov 04, 2019 11:33 pm
Forum: Lewis Structures
Topic: Exceptions for the octet rule
Replies: 5
Views: 128

### Re: Exceptions for the octet rule

Many of the elements with n is greater or equal to 3 can have an expanded octet, for example, chlorine and iodine. In addition, Boron and Aluminum sometimes do not have a complete octet and instead only have 6 valence electrons as it minimizes formal charge (ie BF3).
Mon Nov 04, 2019 11:25 pm
Forum: Student Social/Study Group
Topic: HW 2D13
Replies: 1
Views: 64

### Re: HW 2D13

The shorter the bond, the stronger it is. Also, the higher order of the bond, the stronger it is. This means that a double bond, which is stronger than a single bond, is also shorter than it. You should draw out the Lewis structures for the molecules and see what type of bonds they form. Then you ca...
Mon Nov 04, 2019 8:24 pm
Forum: Trends in The Periodic Table
Topic: Effective Nuclear Charge
Replies: 5
Views: 295

### Re: Effective Nuclear Charge

The s orbitals are closer to the nucleus that the p orbitals, so there is shielding of the electrons in the p orbital by the ones in the s orbital. This means the outermost electrons will experience less nuclear attraction while the inner ones experience greater attraction because of the increase in...
Tue Oct 29, 2019 12:03 pm
Forum: Ionic & Covalent Bonds
Topic: Polar vs Nonpolar bond strength
Replies: 2
Views: 57

### Polar vs Nonpolar bond strength

Does the polarity of bonds have any effect on their strength? For example, are polar or nonpolar bonds stronger?
Tue Oct 29, 2019 11:50 am
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Excited vs. Ground State Configurations 1E.7
Replies: 5
Views: 80

### Re: Excited vs. Ground State Configurations 1E.7

There is only one ground state for an atom (when it has the lowest potential energy). You find this configuration by following the various rules. Aufbau's rule states that you fill the lowest energy 1s 2s 2p ect orbitals first. Hund's rule states that you should have as many electrons with parallel...
Tue Oct 29, 2019 11:37 am
Forum: Formal Charge and Oxidation Numbers
Topic: Formal Charge Equation
Replies: 2
Views: 54

### Re: Formal Charge Equation

The variables may vary, but the general form of the equation is FC (formal charge) = V (valence electrons) - L (number of lone pair electrons) -1/2 B (half of bonding electrons) For example, consider CO2. Looking at the carbon, you can see it should have 4 valence electrons. The lewis dot diagram sh...
Tue Oct 29, 2019 11:28 am
Forum: Ionic & Covalent Bonds
Topic: Homework Question 2B3
Replies: 1
Views: 33

### Re: Homework Question 2B3

First, you have to determine the least electronegative element as your center atom, in this case Bromine. You then add the 3 Fluorine with single bonds. Then add up all the valence electrons, in this case 28, around the molecule. Now, the goal is to minimize the formal charge on the molecule. Typica...
Tue Oct 29, 2019 11:18 am
Forum: Bond Lengths & Energies
Topic: bond lengths
Replies: 10
Views: 272

### Re: bond lengths

The more electrons that interact with the bond, the stronger the attraction of the electrons to the nucleus, meaning that the 2 nuclei get pulled closer together and are shorter.
Mon Oct 28, 2019 10:53 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Excited vs. Ground State Configurations 1E.7
Replies: 5
Views: 80

### Re: Excited vs. Ground State Configurations 1E.7

There is only one ground state for an atom (when it has the lowest potential energy). You find this configuration by following the various rules. Aufbau's rule states that you fill the lowest energy 1s 2s 2p ect orbitals first. Hund's rule states that you should have as many electrons with parallel ...
Wed Oct 23, 2019 10:59 am
Forum: Electron Configurations for Multi-Electron Atoms
Topic: 1E.13
Replies: 2
Views: 30

### Re: 1E.13

The reasoning for silver's different electron configuration is the same as the reasoning for copper and chromium. This is an exception to the rule, as having a half filled d orbital will cause the overall potential of the atom to be less. So for the case where the d orbital has 4 or 9 electrons, usu...
Tue Oct 22, 2019 11:33 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: 4s and 3d
Replies: 5
Views: 80

### Re: 4s and 3d

For n<20 4s comes before 3d, as when the 3d orbital is empty, the energy of the orbital is higher. The reasoning of why the 2 switch when n>20 is becuase protons are added to the nucleus and electrons surrounding the nucleus. The various attractions and repulsions in the atoms cause the energies to ...
Tue Oct 22, 2019 11:23 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: HW Question 1E5 d
Replies: 1
Views: 52

### Re: HW Question 1E5 d

Zeff for the s orbital is greater than the p orbital. Zeff is the effective nuclear charge on the electrons. This is affected by the number of protons, or positive charge in the nucleus as well as any shielding that may occur. The s orbital has a probability density that is closer to the nucleus so ...
Tue Oct 22, 2019 5:38 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Excited state vs ground state
Replies: 4
Views: 54

### Re: Excited state vs ground state

The ground state of an electron is the configuration of electrons such that the atom is at the lowest potential energy. You can determine this by following the examples done in class and following Hund's, Aufbau's, and the Pauli Exclusion Principle to keep adding electrons to the lowest energy orbit...
Mon Oct 21, 2019 9:50 pm
Forum: Properties of Electrons
Topic: Diffraction Patterns for Electrons
Replies: 3
Views: 38

### Re: Diffraction Patterns for Electrons

Yes that's right. Using the deBroglie equation, you can see that that the wavelength of the electron is very noticeable compared to its size, compared to a baseball, whose wavelength is negligible and undetectable. All waves will have diffraction patterns, so when the electron is acting as a wave, t...
Wed Oct 16, 2019 6:42 pm
Forum: Quantum Numbers and The H-Atom
Topic: Homework 1B25 Help
Replies: 1
Views: 62

### Re: Homework 1B25 Help

To solve this problem, you need to use Heisenberg's Equation, which is delta p * delta x >= h/(4pi), where h is a Plank's constant, delta x is the uncertainty of position, and delta p is uncertainty in momentum. The goal of the question is to find the minimum uncertainty in speed which you can find ...
Wed Oct 16, 2019 4:34 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: 1D.25
Replies: 3
Views: 67

### Re: 1D.25

Even though the periodic table ends at 4f and 5f, theoretically there could exist an atom if it had more mass that would have a 6f orbital. This element is either undiscovered yet or highly unstable or both. What the question is asking is more theoretical, and looking for what subshells l can exist ...
Wed Oct 16, 2019 4:27 pm
Forum: Trends in The Periodic Table
Replies: 11
Views: 98

As you go down a group, the principle quantum number, n, increases, which means that the distance from the nucleus of the outermost valance electrons also increases. This is also the case for atomic radii as well.
Tue Oct 15, 2019 1:24 am
Forum: *Shrodinger Equation
Topic: Wave Function
Replies: 2
Views: 76

### Re: Wave Function

When he says psi function, he means a mathematical way to describe the position of the electron. If you simplify it to 2D, one way to describe a position of the electron as a wave is to use a sin or cos function. The representation of this would be the graph of the wave. For the 1s orbital, ψ=1/√(π)...
Tue Oct 15, 2019 1:14 am
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Module Question 29
Replies: 1
Views: 41

### Re: Module Question 29

You should use E = hv to find the energy per photon. After calculating that, you can find the number of photons by dividing total energy (11 J) by the energy per photon.
Fri Oct 11, 2019 12:59 am
Forum: Quantum Numbers and The H-Atom
Topic: Lines in the emission spectra
Replies: 1
Views: 46

### Re: Lines in the emission spectra

The entire graph gives us a range of different wavelengths, from 0 to a large number. We define what each range of the spectrum is called. For example, visible light is in the region between 400 and 700 nm. Infrared radiation, which has a longer wavelength may be found directly next to the visible l...
Fri Oct 11, 2019 12:52 am
Forum: Quantum Numbers and The H-Atom
Topic: Energy quantized?
Replies: 5
Views: 103

### Re: Energy quantized?

Quantized means that the amount of energy released is discrete, not continuous. The values that the energy can take are limited and only certain ones are allowed. This means that an electron might only be able to release only 4.5 * 10^-19 joule or 1.3 * 10^-18 joules of energy but not 8.3 *10 ^-19 o...
Fri Oct 11, 2019 12:26 am
Forum: Properties of Electrons
Topic: Problem B.15
Replies: 2
Views: 57

### Re: Problem B.15

You can use the deBroglie relation lambda = h/p. To find the wavelength, you know h the constant, and can find momentum p by multiplying mass times velocity of the electron.
Wed Oct 09, 2019 12:46 am
Forum: Accuracy, Precision, Mole, Other Definitions
Topic: Mole help [ENDORSED]
Replies: 8
Views: 160

### Re: Mole help[ENDORSED]

Yes that's right. You can check if you're doing the right thing by writing out the dimensional analysis. Start with grams and multiply by mols/gram. Just like multiplying fractions, the grams in both cancel out and you are left with moles. To convert from moles to grams, you would instead multiply b...
Wed Oct 09, 2019 12:40 am
Forum: Molarity, Solutions, Dilutions
Topic: Question L37, Volume of H2O
Replies: 2
Views: 63

### Re: Question L37, Volume of H2O

I think 800 mL is just an approximate amount of water so that your final volume is 1 L. You always want to add acid to water for safety and then slowly adjust the solution with more water until it reaches the 1L mark. It could have been 700 mL or 900 mL or anything around there.
Sat Oct 05, 2019 2:01 am
Forum: Molarity, Solutions, Dilutions
Topic: G 25.
Replies: 1
Views: 69

### Re: G 25.

You don't really need to compute that for the question, as you just need to consider the number of atoms of substance X in the final solution (the molarity of the solution is (1/2)^90 of the original). However, to find how many times you would need to dilute it to reach 1 atom, you would calculate t...
Fri Oct 04, 2019 5:45 pm
Forum: Properties of Light
Topic: Problem 1A.15 and the Rydberg formula
Replies: 3
Views: 53

### Re: Problem 1A.15 and the Rydberg formula

The method I use is a guess and check. If you divide the 2 numbers, you get that (1/n12 - 1/n22) = .8875. You know that the final energy level cannot be greater than 1 as if n is 2, then 1/(n^2) will be .25 which is much less than .8875. I try to approximate the decimal as a fraction, in this case ....
Fri Oct 04, 2019 5:28 pm
Forum: Limiting Reactant Calculations
Topic: Homework M.15
Replies: 2
Views: 63

### Re: Homework M.15

This is a limiting reactants problem and to solve this, you need to see which of the reactants, Al or Cl, will be used up first. To do this, you must calculate how many moles of each reactant there are and how many moles of product they can produce. You can multiply the mass of each by the molar mas...