Search found 49 matches

by Julian Krzysiak 2K
Sat Mar 17, 2018 9:22 pm
Forum: General Rate Laws
Topic: Test 3 Q3- Friday discussion
Replies: 1
Views: 278

Re: Test 3 Q3- Friday discussion

Yup, that's the right answer.
by Julian Krzysiak 2K
Sat Mar 17, 2018 8:57 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Electron Transfer from Oxidation Number
Replies: 1
Views: 103

Re: Electron Transfer from Oxidation Number

From what I can gather, the only molecule in this reaction that has an overall charge is 02. Each one is -2, so overall it is -4, and because there are 6 02, there is a charge of -24 on it. Only the reactant side has a a negative charge, so I would put 24 electrons on the other side to balance out t...
by Julian Krzysiak 2K
Sat Mar 17, 2018 8:48 pm
Forum: Balancing Redox Reactions
Topic: Notation question
Replies: 2
Views: 196

Re: Notation question

Apparently NaN3 is an ionic compound, so the Na and N3 are attracted by their difference in charges. Na has an oxidation state of +1, but with the N3, we have to figure out the overall charge of each nitrogen via formal charges. How they have it N=N=N, the outermost Nitrogens have a formal charge of...
by Julian Krzysiak 2K
Sat Mar 17, 2018 8:19 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Catalysts
Replies: 3
Views: 224

Re: Catalysts

We know if there is a catalyst if it was present in the reactants in the first step, and is present as a product in the last step. It is never consumed. This is opposed to an intermediate, which is formed in the first step and later consumed in a later step. A+B->C C+D->E+B A+D->E In this, B is the ...
by Julian Krzysiak 2K
Sat Mar 17, 2018 8:12 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 14.15a
Replies: 2
Views: 115

Re: 14.15a

I believe it is because the Ag(s) is acting as an electrode. Although they are both solids, it would be difficult to tell which is the electrode if you just put a comma as they are both solids, so they put a line to differentiate between the electrode and the "solution".
by Julian Krzysiak 2K
Sat Mar 17, 2018 8:04 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Entropy change in volume or pressure
Replies: 1
Views: 203

Re: Entropy change in volume or pressure

For both equations, the temperature is not changing, isothermal. The volume and pressure for both equations aren't constant, due to PV=nRT. The pressure and volume are inversely proportional, so changing one changes the other.
by Julian Krzysiak 2K
Wed Mar 14, 2018 2:44 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Can someone help me identify the oxidizing and reducing agents in this galvanic cell?
Replies: 1
Views: 135

Re: Can someone help me identify the oxidizing and reducing agents in this galvanic cell?

O_{3} is the oxidizing agent O_{2} is the reducing agent I found this out by "matching" what we were given in the appendix of reducing half-reactions given to us for the test. For O3/O2, OH- , the standard reduction potential of O3 + H20 + 2e -> O2 + 2OH matches For O3, H+/O2, the standar...
by Julian Krzysiak 2K
Wed Mar 14, 2018 2:18 am
Forum: Phase Changes & Related Calculations
Topic: Peer Learning Prob on heat
Replies: 1
Views: 106

Re: Peer Learning Prob on heat

By setting both heats opposite to each other, did you mean making them both equal to each other?
If so, you'd need to also set one side as negative, because
by Julian Krzysiak 2K
Wed Mar 14, 2018 2:10 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Test 2 Question 5
Replies: 2
Views: 122

Re: Test 2 Question 5

Your logic is correct, Mn has the most reducing power. But the question asks to put them in order of increasing reducing power, which means you have to order them from worst reducing power to best reducing power. The way you wrote it is technically correct, with the inequality signs, Mn>Zn>Cr. If yo...
by Julian Krzysiak 2K
Mon Mar 12, 2018 12:33 am
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Example 15.7
Replies: 2
Views: 97

Re: Example 15.7

I doubt it, since the method requires a graphing calculator, and we haven't gone over that method in class.
by Julian Krzysiak 2K
Mon Mar 12, 2018 12:31 am
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Pre-equilibrium Condition
Replies: 3
Views: 151

Re: Pre-equilibrium Condition

It's saying that when the first step is fast, and the second step is slow, then when the first step creates the intermediate, it does this at a much faster rate than the second step can consume it. So a bottleneck is created, where you have a bunch of intermediate waiting to be consumed in the slow ...
by Julian Krzysiak 2K
Mon Mar 12, 2018 12:16 am
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Collision Cross Section
Replies: 2
Views: 155

Re: Collision Cross Section

The collision cross section is important because it makes up a component of A, the pre-exponential factor. A=\sigma \bar{v} {N_{a}}^{2} The pre-exponential factor tells us the rate at which molecules collide with each other. The collision cross section plays into this because it gives us the area of...
by Julian Krzysiak 2K
Sun Mar 11, 2018 11:58 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Steric requirement
Replies: 1
Views: 128

Re: Steric requirement

When calculating the rate constant using the collision theory, we need to take into account the orientation of the colliding atoms/molecules. This is the steric requirement, P. This steric factor is usually always less than 1 because the steric requirement lowers the probability of the reaction occu...
by Julian Krzysiak 2K
Mon Mar 05, 2018 12:51 am
Forum: General Rate Laws
Topic: Homework Problem 15.19C
Replies: 5
Views: 184

Re: Homework Problem 15.19C

I was able to get 2.85x10^(12) by converting everything from mmol into mol. By dividing everything by 1000 first, the initial reactant concentrations and initial rate, you get k=2.85x10^(12). Only problem is that the book states the answer in mmol still, so this may possibly be an error in the answe...
by Julian Krzysiak 2K
Mon Mar 05, 2018 12:38 am
Forum: First Order Reactions
Topic: Half-life of first order reactions
Replies: 3
Views: 134

Re: Half-life of first order reactions

Yup, that is exactly correct!

From the book, "The half-life of a reactant that decays in a first-order reaction is characteristic of the reaction and independent of the initial concentration. It is inversely proportional to the rate constant of the reaction."
by Julian Krzysiak 2K
Sun Feb 18, 2018 9:37 pm
Forum: Balancing Redox Reactions
Topic: Basic v Acidic
Replies: 2
Views: 97

Re: Basic v Acidic

They will always tell us whether the reaction is basic or acidic in the question, and from that knowledge, we will know when to use H+ (acidic) or OH- (basic) to balance out the reaction. For either one, though, you'll have to use H20 to balance out the reaction.
by Julian Krzysiak 2K
Sun Feb 11, 2018 10:15 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Problem 9.75 Clarification
Replies: 1
Views: 87

Re: Problem 9.75 Clarification

Residual Entropy is the entropy entirely dependent on the positional disorder when T=0. All entropy still available can't be due to the Temperature, but there is still an inherent entropy due to the different microstates of the substance. A trans-isomer is pretty much an isomer where 2 parts are on ...
by Julian Krzysiak 2K
Sun Feb 11, 2018 9:42 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Sublimation
Replies: 3
Views: 209

Re: Sublimation

The only places where I've seen Enthalpy of Sublimation for a specific substance used is when we are using mean bond enthalpies to calculate the enthalpy for the reaction. We would use sublimation when we need to create a new molecule in a gaseous form, but one of the reactants is in a solid state. ...
by Julian Krzysiak 2K
Sun Feb 11, 2018 9:32 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: 9.43
Replies: 1
Views: 126

Re: 9.43

We can't use that equation because we don't know the final temperature, we are only given the initial temperatures of both bodies of water. That means we can use this equation, \Delta S=nCln\frac{T_{1}}{T_{2}} , if we know the final temperature, because then we would know the final temperatures of b...
by Julian Krzysiak 2K
Mon Feb 05, 2018 12:20 am
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Question: Refrigerator Cooling
Replies: 3
Views: 145

Re: Question: Refrigerator Cooling

I think it's only because the surroundings in this case are so much more larger and have much more volume as compared to the refrigerator and what it produces. To cool the whole room down would require a large amount of energy, which would be theoretically possible if the refrigerator were large eno...
by Julian Krzysiak 2K
Mon Feb 05, 2018 12:16 am
Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
Topic: 9.43
Replies: 2
Views: 139

Re: 9.43

Because we need to know the entropy of both the system (20 C) and the surroundings (50 C), we need to use the equation for finding the entropy with the final and initial temperatures. Unfortunately, they don't give us the final Temperature, so we have to find it ourselves. So you set both of the equ...
by Julian Krzysiak 2K
Mon Feb 05, 2018 12:09 am
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: 9.19 Finding standard entropy of water
Replies: 1
Views: 83

Re: 9.19 Finding standard entropy of water

Because Entropy is a state function, we can take any amount of paths to get to the final value. Because we have water going from a liquid to a gas, we have to take into account the phase change, so we need to add the equations of the liquid reaching the boiling point and its vaporization. From the g...
by Julian Krzysiak 2K
Sun Jan 28, 2018 10:53 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Test question [ENDORSED]
Replies: 16
Views: 565

Re: Test question [ENDORSED]

I put that the First Law of Thermodynamics states that the internal energy of an isolated system will not change. I also put just in case. Hopefully that's specific enough.
by Julian Krzysiak 2K
Sun Jan 28, 2018 10:50 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Heat capacity [ENDORSED]
Replies: 4
Views: 145

Re: Heat capacity [ENDORSED]

No, any heat capacities, along with any relevant constants/equations, will be provided to us with every test on an equations sheet.
by Julian Krzysiak 2K
Sun Jan 28, 2018 10:24 pm
Forum: Phase Changes & Related Calculations
Topic: degeneracy [ENDORSED]
Replies: 7
Views: 195

Re: degeneracy [ENDORSED]

Degeneracy is the total number of ways of achieving a given energy state. In the dogbone example Dr. Lavelle showed us, we could could calculate W with the info given by W_{N}=2^{N} . N is the number of particles, molecules, etc., while the 2 was the number of equivalent states. So if there was 1 pa...
by Julian Krzysiak 2K
Sun Jan 21, 2018 6:18 pm
Forum: Phase Changes & Related Calculations
Topic: 8.39
Replies: 1
Views: 95

Re: 8.39

Yes, since the solid is already at its melting point, you can start with the phase change calculation, and then add that to the heat needed to get the water to the specific temperature.
by Julian Krzysiak 2K
Sun Jan 21, 2018 6:13 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Cv vs. Cp [ENDORSED]
Replies: 13
Views: 956

Re: Cv vs. Cp [ENDORSED]

Cp is the heat capacity for something at a constant pressure, and would be equal to \Delta Cp=\Delta H/\Delta T Cv is the heat capacity for something at a constant volume, and would be equal to \Delta Cv=\Delta U/\Delta T You'll know when to use a particular heat capacity because they'll specify in ...
by Julian Krzysiak 2K
Sun Jan 21, 2018 5:54 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: 8.25
Replies: 3
Views: 186

Re: 8.25

The regular heat equation for any system/surroundings is Qsystem=-Qsurroundings, because the amount of energy must be conserved between the 2 environments, due to the Law of the Conservation of Energy. In this case, the reaction=system, and surroundings=calorimeter, so Qreaction= -Qcalorimeter. Then...
by Julian Krzysiak 2K
Sun Jan 14, 2018 10:38 pm
Forum: Phase Changes & Related Calculations
Topic: Heat Capacity
Replies: 4
Views: 174

Re: Heat Capacity

And extensive property is a property that changes based on how much there is of the sample- it is based on weight, for example, mass, volume, etc. An intensive property is a property that is an intrinsic quality of the sample- it does not change based on weight, for example, density, hardness, etc. ...
by Julian Krzysiak 2K
Sun Jan 14, 2018 10:37 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Extensive vs Intensive Properties
Replies: 3
Views: 139

Re: Extensive vs Intensive Properties

And extensive property is a property that changes based on how much there is of the sample- it is based on weight, for example, mass, volume, etc. An intensive property is a property that is an intrinsic quality of the sample- it does not change based on the weight, for example, density, hardness, e...
by Julian Krzysiak 2K
Sun Jan 14, 2018 10:24 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Standard State
Replies: 2
Views: 124

Re: Standard State

Elements are in their standard state when they are in their most stable phase at the temperature of interest, which is usually 25 Celsius, while exposed to a constant pressure of 1 atm (They must also be 1 mole if they are a solution). So you can look at a periodic table at 25 C. and check the phase...
by Julian Krzysiak 2K
Sun Dec 03, 2017 10:54 pm
Forum: Shape, Structure, Coordination Number, Ligands
Topic: Oxidation States
Replies: 4
Views: 190

Re: Oxidation States

Are you saying if you want to find the oxidation state for anything, or just the transition metal in the coordination complex? For many elements, you can find the oxidation state by imagining what would happen if the element were in an ionic bond, and were to have a full octet. How many electrons wo...
by Julian Krzysiak 2K
Sun Dec 03, 2017 10:39 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: Ph, Poh, and kw
Replies: 2
Views: 195

Re: Ph, Poh, and kw

The Kw constant is another name for the autoprotolysis constant. Autoprotolysis is a reaction where water naturally decomposes into hydronium and hydroxide, due to the amphiprotic nature of water allowing proton transfer between the same type of molecule. The equilibrium constant would be found like...
by Julian Krzysiak 2K
Sun Dec 03, 2017 10:15 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 11.47
Replies: 1
Views: 146

Re: 11.47

Because we are given grams of the reactant, we have to convert to concentration by converting the grams to moles, and then dividing that by the vessel volume. 1.0g/208.22 moles --> .00480 mole/.250 Liters ---> .0192 mol./L. Then we would insert this concentration (.0192) into the ICE box, in the &qu...
by Julian Krzysiak 2K
Sun Dec 03, 2017 9:56 pm
Forum: Hybridization
Topic: Pi-Bonds and their interactions
Replies: 2
Views: 166

Re: Pi-Bonds and their interactions

A sigma bond has overlapping orbitals in a straight line, while pi bonds have orbitals that overlap when the p orbitals are parallel to each other. When you have a double bond, it's a sigma bond, along with the other p orbitals oriented so that they can also create a bond. The other post explained i...
by Julian Krzysiak 2K
Sun Nov 26, 2017 5:08 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Equilibrium Constant units
Replies: 2
Views: 154

Re: Equilibrium Constant units

K is a ratio between the concentrations/partial pressures of the reactants and products when they have reached equilibrium. The units of the products and reactants would cancel each other out. Also, K being a ratio would mean its just a numerical relationship between two different concentration/part...
by Julian Krzysiak 2K
Sun Nov 26, 2017 4:55 pm
Forum: Shape, Structure, Coordination Number, Ligands
Topic: Some clarification about ligands
Replies: 5
Views: 209

Re: Some clarification about ligands

Ligand is another name for a Lewis base which bonds to a central metal atom/ion. A Lewis base is able to do this by the presence of lone pairs, allowing the atom to "donate" its electrons and covalently bond to the central metal atom. This bond is called a coordination covalent bond, and t...
by Julian Krzysiak 2K
Sun Nov 19, 2017 11:58 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: seesaw shape
Replies: 1
Views: 168

Re: seesaw shape

The see-saw shape is based off of the trigonal bipyramidal, as the molecule in question has 5 areas of electron density. But then you would just need to replace one of the bonds with a lone pair, and in this case it would be the equatorial bond, as you would only be strongly repelling against 2 axia...
by Julian Krzysiak 2K
Sun Nov 19, 2017 10:18 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Angular
Replies: 2
Views: 126

Re: Angular

Yes, they're the same thing.

And for 4.5 a, it would be angular (bent) since the central Cl atom has a double bond with each O atom along with a lone pair. This lone pair repels the other bonds as far away as possible to attain the lowest energy, resulting in the bent shape similar to H20.
by Julian Krzysiak 2K
Sun Nov 05, 2017 7:57 pm
Forum: Lewis Structures
Topic: HW 3.59 PART C
Replies: 1
Views: 91

Re: HW 3.59 PART C

If you drew N with double bonds to both O, then you would have 10 electrons for N, since you also have to draw a single bond from N to the other O that has Cl attached. You can only have one double bond with 1 O, because then you would be able to have 4 bonds connecting to N, fulfilling the octet ru...
by Julian Krzysiak 2K
Sun Nov 05, 2017 7:48 pm
Forum: Formal Charge and Oxidation Numbers
Topic: 3.49
Replies: 3
Views: 253

Re: 3.49

When solving for formal charge, the L is the # of electrons in the bond pair, so it would be 2; you don't count the whole pair as 1.

So if we do that, then O would be 6 -(2+ 6/2)= +1, and N would be 5 -(2 + 6/2)= 0
by Julian Krzysiak 2K
Sun Oct 29, 2017 11:36 pm
Forum: Lewis Structures
Topic: Lewis Structure and Resonance
Replies: 1
Views: 78

Re: Lewis Structure and Resonance

While resonance does have to do with the possibility of multiple bonds, it is important to match up the correct number of e- with each resonance bond. If you are talking about a resonance hybrid, when you combine all the resonance structures, you still have to include the matching number of lone pai...
by Julian Krzysiak 2K
Sun Oct 29, 2017 11:04 pm
Forum: Resonance Structures
Topic: Resonance Structures [ENDORSED]
Replies: 3
Views: 239

Re: Resonance Structures [ENDORSED]

A resonance structure would just be the Lewis structure of all the possible configurations of the bonds in a molecule. This is signified by a double arrow between each possible Lewis structure, showing the possible resonance structures of the molecule. But the best representation of a molecule would...
by Julian Krzysiak 2K
Sun Oct 22, 2017 6:07 pm
Forum: DeBroglie Equation
Topic: Exercise 1.42 [ENDORSED]
Replies: 2
Views: 180

Re: Exercise 1.42 [ENDORSED]

I would assume you need to use the de Broglie Equation, \lambda =h/m_{h}*v They give us the velocity, and we already know what constant h is, so we would just need to know the mass of a helium atom. We would get that by dividing the molar mass of Helium, 4.0026, by Avogadro's constant, 6.02214*10^{2...
by Julian Krzysiak 2K
Tue Oct 17, 2017 6:18 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Question 1.15
Replies: 6
Views: 266

Re: Question 1.15

v=R(1/n_{1}^{2}-1/n_{2}^{2}) is just another way of writing \Delta E= E_{final}- E_{initial} When we convert the \Delta E equation, the E_{final} corresponds to n_{1} , and E_{initial} corresponds to n_{2} We know that n_{1}=1 , the final energy level, and n_{2}=3 , the initial energy level...
by Julian Krzysiak 2K
Sun Oct 15, 2017 5:07 pm
Forum: Properties of Electrons
Topic: Atomic Spectra and Energy Levels
Replies: 3
Views: 219

Re: Atomic Spectra and Energy Levels

The equation for the energy of an electron at any energy level is

From this equation, we know that as n goes higher, the energy will keep getting smaller. When n is equal to infinity, the Energy will be equal to 0, since dividing by infinity results in 0.
by Julian Krzysiak 2K
Sun Oct 15, 2017 4:44 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Variation of Empirical Equation of H-atom [ENDORSED]
Replies: 2
Views: 171

Re: Variation of Empirical Equation of H-atom [ENDORSED]

\Delta E = E_{final}-E_{initial} so we can plug in E_{n}=-hR/n^2 for the final and initial eqation, \Delta E= (-hR/n_{1}^2) - (-hR/n_{2}^2) We can make E_{f} have n_{1} because it is the final and lower energy level, which is the case when the electron is moving down an energy level...
by Julian Krzysiak 2K
Sun Oct 08, 2017 10:55 pm
Forum: Student Social/Study Group
Topic: Post All Chemistry Jokes Here
Replies: 7348
Views: 890794

Re: Post All Chemistry Jokes Here

All these jokes were noble attempts, yet I didn't really react to any of them.
by Julian Krzysiak 2K
Sun Oct 08, 2017 10:25 pm
Forum: Student Social/Study Group
Topic: Post All Chemistry Jokes Here
Replies: 7348
Views: 890794

Re: Post All Chemistry Jokes Here

What's a graduate student's favorite element?

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