Search found 54 matches

by Seth_Evasco1L
Fri Mar 16, 2018 8:27 pm
Forum: Balancing Redox Reactions
Topic: Oxidation state
Replies: 7
Views: 285

Re: Oxidation state

What would the oxidation state of I3- be?
by Seth_Evasco1L
Fri Mar 16, 2018 8:24 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: Michaelis-Menten rate equation
Replies: 2
Views: 112

Re: Michaelis-Menten rate equation

This was covered in another post and since it was not covered in class, then this won't be covered on the final.
by Seth_Evasco1L
Fri Mar 16, 2018 3:14 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 14.13 B
Replies: 3
Views: 149

Re: 14.13 B

Solids can be used as electrodes but only if they are able to conduct electricity. Iodine (s), being a halogen, is a nonmetal and is therefore unable to conduct electricity. This is why a platinum electrode is utilized at the anode to act as an inert metal that will assist in conducting electricity.
by Seth_Evasco1L
Thu Mar 15, 2018 7:24 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Heat curve
Replies: 2
Views: 232

Re: Heat curve

Yes that would be correct! If we look at a heating curve, the x-axis would be the heat absorbed and the y-axis would be the temperature. Let's say for some substance, the heat capacity of a substance in a solid phase is 10 J/ o C and the heat capacity of the same substance is 20 J/ o C in its liquid...
by Seth_Evasco1L
Thu Mar 15, 2018 3:44 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Question 15.51
Replies: 8
Views: 311

Re: Question 15.51

Yes, that would be the bottleneck concept where the intermediate is built up due to the fact that a fast step occurs before the slow step. In this case, you would use the pre-equilibrium approximation approach.
by Seth_Evasco1L
Thu Mar 15, 2018 2:04 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: endothermic reaction.
Replies: 5
Views: 231

Re: endothermic reaction.

For example, there's an endothermic reaction plotted on a graph where Gibbs free energy (G) is plotted against the progress of the reaction. The reactants would be at a lower G than the products. Thus, the activation energy for the forward reaction would be higher in the forward direction than in th...
by Seth_Evasco1L
Thu Mar 15, 2018 2:00 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: k' reverse reaction constant
Replies: 6
Views: 312

Re: k' reverse reaction constant

Say for example A + B --> C + D The forward rate can be represented as k[A][B] and the reverse rate can be represented as k'[C][D]. At equilibrium you can set these rates equal to each other to get: k[A][B] = k'[C][D] This can be rewritten as: k/k' = [C][D]/[A][B] K, the equilibrium constant = [prod...
by Seth_Evasco1L
Thu Mar 15, 2018 1:46 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: Test #2 Question 8d
Replies: 1
Views: 113

Re: Test #2 Question 8d

E = Eo - ((RT/nF) * ln(Q))

Plug in the corresponding values to get:

E = 0.37 - ((8.314*298/1*96485) * ln(26.8))

E = 0.29V
by Seth_Evasco1L
Thu Mar 15, 2018 10:50 am
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Slow vs Fast Reactions
Replies: 2
Views: 140

Re: Slow vs Fast Reactions

In order to determine which step is the slow or fast step, we must be given the experimentally determined rate law in order to test which step matches the proposed mechanism.
by Seth_Evasco1L
Thu Mar 15, 2018 10:49 am
Forum: Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy
Topic: NO vs BF3
Replies: 1
Views: 225

Re: NO vs BF3

NO has resonance due to the lone pair that can be positioned either on the Oxygen or Nitrogen atom. It has more orientation positions than BF 3 which does not have resonance since Boron will have a single bond with each Fluorine atom with no leftover electrons. This means that the degeneracy of NO w...
by Seth_Evasco1L
Thu Mar 15, 2018 10:32 am
Forum: Second Order Reactions
Topic: Half-Life of a Second Order Reaction
Replies: 3
Views: 295

Re: Half-Life of a Second Order Reaction

Thank you so much!
by Seth_Evasco1L
Thu Mar 15, 2018 10:26 am
Forum: Second Order Reactions
Topic: Half-Life of a Second Order Reaction
Replies: 3
Views: 295

Half-Life of a Second Order Reaction

Can someone please explain how the half-life of a second order reaction works compared to a first order reaction?
by Seth_Evasco1L
Thu Mar 15, 2018 10:24 am
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Reaction Mechanisms
Replies: 2
Views: 125

Re: Reaction Mechanisms

1. Direct Computation by solving differential equations (won't need to know this) 2. Identify rate-limiting step. Use rate law of rate-limiting step as overall rate. Compare model w/ experiment. a. Steady State Approximation: constant intermediate concentration in rate-limiting step (won't need to k...
by Seth_Evasco1L
Thu Mar 15, 2018 10:11 am
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Standard Gibbs
Replies: 3
Views: 352

Re: Standard Gibbs

Standard just indicates that all reactants and products are in their standard state.

Standard state for a gas is 1 atm. Standard state for a solution is 1M at 1 atm. Standard state for an element is the most stable phase (solid, liquid, gas) at 1 atm and temperature of interest (250C).
by Seth_Evasco1L
Tue Mar 13, 2018 4:21 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: HW 15.49
Replies: 1
Views: 104

Re: HW 15.49

We can only raise concentrations to their respective coefficients in elementary steps such as those proposed in the reaction mechanism given. The mechanism is proposing in 15.49 for instance that one molecule of HBr collides with one molecule of NO2 in step 1 and similarly in step 2.
by Seth_Evasco1L
Tue Mar 13, 2018 11:38 am
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Specific Heat Capacity vs. deltaHfus and deltaHvap
Replies: 2
Views: 253

Re: Specific Heat Capacity vs. deltaHfus and deltaHvap

Adding onto the previous post, since those three terms don't depend on the amount of substance into account, they can be described as intensive properties.
by Seth_Evasco1L
Tue Mar 13, 2018 1:35 am
Forum: Biological Examples
Topic: Nitric Oxide Catalyses
Replies: 6
Views: 514

Re: Nitric Oxide Catalyses

Species that appear in the overall reaction can be ruled out as being the intermediate or catalyst. An intermediate is produced in one elementary step and then consumed in a subsequent elementary step. A catalyst is found on the reactants side of one elementary step and also on the products side in ...
by Seth_Evasco1L
Tue Mar 13, 2018 1:18 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Cell Diagrams- Line Notation
Replies: 4
Views: 191

Re: Cell Diagrams- Line Notation

Additionally, those single lines should only be used if there is a difference in phase between the reactants and products of the anode/cathode. Otherwise, use a comma to separate them.
by Seth_Evasco1L
Tue Mar 13, 2018 12:51 am
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: Problem 15.61
Replies: 5
Views: 212

Re: Problem 15.61

I calculated using joules and had my answer off by a factor of 103 which is reasonable because converting to kilojoules would provide me with the same answer as the solution manual. Perhaps there was some miscalculation in the process?
by Seth_Evasco1L
Tue Mar 13, 2018 12:45 am
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: 15.63
Replies: 3
Views: 188

Re: 15.63

Yes! That should be a typo as 37oC should be 310 K when converted over.
by Seth_Evasco1L
Tue Mar 13, 2018 12:44 am
Forum: Reaction Mechanisms, Reaction Profiles
Topic: 15.63
Replies: 5
Views: 202

Re: 15.63

Another typo is in the initial statement at the top of the solution given. The temperature in Kelvin should actually read 310 K and not 3130 K.
by Seth_Evasco1L
Tue Mar 13, 2018 12:43 am
Forum: Reaction Mechanisms, Reaction Profiles
Topic: 15.63
Replies: 5
Views: 202

Re: 15.63

I believe there are several typos in 15.63 of the solution manual: The first typo being that 0.08314 kJ*K -1 *mol -1 should really be 0.008314 kJ*K -1 *mol -1 when converting the rate constant R in terms of kJ. When you use that to calculate ln(k'/k), you should get approximately 0.59 which is where...
by Seth_Evasco1L
Tue Mar 13, 2018 12:31 am
Forum: General Rate Laws
Topic: 15.99 (e)(f)
Replies: 3
Views: 155

Re: 15.99 (e)(f)

For e) Plotting k against temperature will not produce a linear graph. If you look at the Arrhenius equation, you'll find that k and T are not directly proportional. If we linearize the equation by taking the natural log of both sides we'll find that ln(k) and 1/T are directly proportional and will ...
by Seth_Evasco1L
Tue Mar 13, 2018 12:25 am
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Question 15.51
Replies: 8
Views: 311

Re: Question 15.51

If the second elementary reaction is significantly faster than the first, then it has no effect in the overall reaction order. We've discussed that the slow step (in this case, the first elementary reaction) is the one that determines the overall reaction order because the overall reaction cannot pr...
by Seth_Evasco1L
Tue Mar 13, 2018 12:11 am
Forum: *Organic Reaction Mechanisms in General
Topic: Functional Groups
Replies: 7
Views: 659

Re: Functional Groups

Yes, functional groups will be on the final. We'll most likely have to name the functional group based on a given molecular formula or structure.
by Seth_Evasco1L
Fri Mar 09, 2018 3:13 pm
Forum: First Order Reactions
Topic: half-lives of first order versus second order
Replies: 3
Views: 258

Re: half-lives of first order versus second order

I believe you're able to use your method for the first order reaction because the half life equation doesn't involve the initial concentration of the reactant, so you could use any value for the initial concentration just as long as you have the concentration at time t=1/2 as 1/8 of that initial val...
by Seth_Evasco1L
Fri Mar 09, 2018 3:00 pm
Forum: General Rate Laws
Topic: Rate Constant k
Replies: 6
Views: 245

Re: Rate Constant k

- As temperature (T) increases, k increases
- As activation energy (Ea) increases, k decreases
- As frequency factor (A) increases, k increases
by Seth_Evasco1L
Fri Mar 09, 2018 2:45 pm
Forum: First Order Reactions
Topic: Calculating t using First Order Integrated Rate Law
Replies: 2
Views: 229

Re: Calculating t using First Order Integrated Rate Law

If the concentration decreased to 50% of its initial concentration, then you can set up the equation like so: ln(0.5(A 0 )) = -k*t 1/2 + ln(A 0 ) You can subtract the ln(A 0 ) term to the left side to get: ln(0.5(A 0 )) - ln(A 0 ) = -k*t 1/2 Using logarithm rules, you can simplify it to: ln(0.5(A 0 ...
by Seth_Evasco1L
Fri Mar 09, 2018 2:37 pm
Forum: Second Order Reactions
Topic: Units [ENDORSED]
Replies: 5
Views: 346

Re: Units [ENDORSED]

Units of k for a reaction order of 1.5 would most likely have units of L0.5*mol-0.5*s-1.
by Seth_Evasco1L
Fri Mar 09, 2018 2:32 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: Order less than 0
Replies: 4
Views: 279

Re: Order less than 0

Also for example, if the rate = k[A]-1 and the concentration of A doubled, then the rate would be halved.
by Seth_Evasco1L
Fri Mar 09, 2018 2:27 pm
Forum: General Rate Laws
Topic: Division in Rate Law
Replies: 3
Views: 163

Re: Division in Rate Law

Yes, it would be zero! Overall order for "rate = k*[A]a/[B]b" is equal to a+b.

For rate = k*[A]/[B], overall order would be a+b or 1+(-1) = 0.
by Seth_Evasco1L
Sun Jan 14, 2018 11:44 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: work vs heat
Replies: 4
Views: 161

Re: work vs heat

I believe that both of those equations are in absence of heat and work respectively. Delta U = q + w Delta U = q when there is no work done on or by the system and there is only heat transfer. Delta U = w when there is no heat transfer and the system has work done on or by it (ex: constant volume in...
by Seth_Evasco1L
Sun Jan 14, 2018 11:36 pm
Forum: Phase Changes & Related Calculations
Topic: Temperature and Heat
Replies: 4
Views: 182

Re: Temperature and Heat

Temperature is typically constant when heat is added to a substance that is undergoing a phase change. Heat energy is stored as potential energy that is used to break the molecular bonds. For example, in ice, as heat is applied at 0 degrees Celsius, the temperature stays constant until the ice has u...
by Seth_Evasco1L
Sun Jan 14, 2018 8:15 pm
Forum: Phase Changes & Related Calculations
Topic: Another 8.41 Question
Replies: 3
Views: 217

Re: Another 8.41 Question

You can think of the ice cube as the system and the water as the surroundings. In relation to the system, the system is gaining heat (positive) and the surroundings are losing heat (negative). The heat transfer is from the surroundings to the system.
by Seth_Evasco1L
Sat Dec 09, 2017 1:43 am
Forum: Ionic & Covalent Bonds
Topic: Ionic Character
Replies: 2
Views: 179

Re: Ionic Character

Remember that electronegativity follows an up and to the right trend. Since O is a period higher than S, we can assume it has a higher electronegativity and thus, it will have more ionic character.
by Seth_Evasco1L
Sat Dec 09, 2017 1:41 am
Forum: DeBroglie Equation
Topic: DeBroglie Equation
Replies: 1
Views: 146

Re: DeBroglie Equation

The DeBroglie equation is calculated by: h/p where h is plank's constant and p is momentum (mass*velocity). Using the information given in the problem, the wavelength is calculated to be 1.4x10 -10 m. The electron in an atom typically has a wavelength of 10 -12 m so since this calculated value is gr...
by Seth_Evasco1L
Sat Dec 09, 2017 1:35 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Chemical Equilibrium Part 2 Module Question #27
Replies: 2
Views: 204

Re: Chemical Equilibrium Part 2 Module Question #27

1. Create an ICE table for the reaction. 2. Find the molarity for H 2 O since it gives you the value in moles. (No need to find the molarity for C because it's a solid and is not involved in the equilibrium constant) 3. The initial concentration for CO and H 2 is going to be 0. 4. You know the equil...
by Seth_Evasco1L
Sat Dec 09, 2017 1:02 am
Forum: Determining Molecular Shape (VSEPR)
Topic: Lone pairs
Replies: 5
Views: 287

Re: Lone pairs

Lone pairs can't directly affect the polarity of a molecule as it doesn't have a dipole moment but it does help influence the shape of the molecule. There are instances where lone pairs can make a molecule polar because it shifts dipole moments so that they are unable to cancel each other out due to...
by Seth_Evasco1L
Sat Dec 09, 2017 12:54 am
Forum: Bond Lengths & Energies
Topic: Bond Length and Bond Strength Relationship
Replies: 5
Views: 555

Re: Bond Length and Bond Strength Relationship

You look at bond length when comparing when there are differing atoms bonded to the Hydrogen atom. For example, there's HCl and HI. The longer the bond length, the stronger the acid. In this case, HI is the stronger acid because its atomic radius creates that longer bond length in comparison to the ...
by Seth_Evasco1L
Sun Dec 03, 2017 9:41 pm
Forum: Hybridization
Topic: Pi-Bonds and their interactions
Replies: 2
Views: 175

Re: Pi-Bonds and their interactions

A double bond is usually consisted of atoms that can perform at least bonds with p-orbitals. The p-orbitals consist of px, py, and pz. These are located on different axes. The first bond that is created is a sigma bond which is when two bonds in the same axis overlap. The pi bond is created in a dou...
by Seth_Evasco1L
Sun Nov 19, 2017 9:57 pm
Forum: Sigma & Pi Bonds
Topic: CHAPTER 4
Replies: 1
Views: 143

Re: CHAPTER 4

Think of sp 3 (tetrahedral - 109.5 degrees) as 25% s-character. Think of sp 2 (trigonal-planar - 120 degrees) as 33% s-character. Using that trend, as s-character increases, the bond angle will increase as well. For example, sp is 50% s-character and has a linear shape which corresponds to 180 degre...
by Seth_Evasco1L
Sun Nov 19, 2017 9:45 pm
Forum: Dipole Moments
Topic: Question 4.29
Replies: 2
Views: 158

Re: Question 4.29

Symmetry is a large factor in deciding which molecules are polar or non-polar. If I remember correctly, Forms 1 and 2 are polar because the bonded Cl atoms are not opposite of each other on the carbon ring. To decide which is more polar, think of the dipole moments created by the C-Cl bonds as a vec...
by Seth_Evasco1L
Fri Nov 17, 2017 9:37 am
Forum: Determining Molecular Shape (VSEPR)
Topic: Electron Density
Replies: 3
Views: 188

Re: Electron Density

The octahedron shape does have 8 faces, but I believe it's the 6 vertices of the shape that we're looking at when we discuss regions of electron density.
by Seth_Evasco1L
Thu Nov 09, 2017 2:53 pm
Forum: Accuracy, Precision, Mole, Other Definitions
Topic: Formula unit
Replies: 1
Views: 186

Re: Formula unit

It's simply a unit of measurement for an ionic compound. For example: - There are 3 atoms of the element, Fluorine. (element - atom) - There are 3 molecules of the molecular compound, CH 4 . (molecular compound - molecule) - There are 3 formula units of the ionic compound, NaCL. (ionic compound - fo...
by Seth_Evasco1L
Wed Nov 01, 2017 7:05 pm
Forum: Quantum Numbers and The H-Atom
Topic: Which Quantum Number Represents What?
Replies: 3
Views: 350

Re: Which Quantum Number Represents What?

n is the principal quantum number: - specifies shell - indicates size l is the orbital angular momentum quantum number: - specifies subshell - indicates shape m l is the magnetic quantum number: - specifies orbitals of the subshell - indicates orientation m s is the magnetic spin quantum number: - s...
by Seth_Evasco1L
Mon Oct 30, 2017 1:26 am
Forum: Quantum Numbers and The H-Atom
Topic: Unpaired electrons (Homework 2.51) [ENDORSED]
Replies: 3
Views: 738

Re: Unpaired electrons (Homework 2.51) [ENDORSED]

Elements that have the 5d orbital can have up to 10 electrons in that specific subshell (I get that it's confusing since that's when the "f" orbitals begin to show up). Tantalum's electron configuration is [Xe]4f 14 5d 3 6s 2 so it would have three unpaired electrons found in the 5d orbital.
by Seth_Evasco1L
Mon Oct 30, 2017 1:03 am
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Question 2.47d [ENDORSED]
Replies: 2
Views: 204

Re: Question 2.47d [ENDORSED]

Gold's electron configuration is [Xe]4f 14 5d 10 6s 1 . When it forms an ion to produce a +1 charge, it will take from the highest energy level, in this case being the 6s orbital. I believe you meant to say the 5d orbital which is typically higher energy than the 6s orbital when we begin filling the...
by Seth_Evasco1L
Sun Oct 29, 2017 3:58 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: +/- x meters ? [ENDORSED]
Replies: 2
Views: 215

Re: +/- x meters ? [ENDORSED]

Yup! That would be correct! If it states "the uncertainty in position is __ meters", then there's no need to double it as that would be the uncertainty value you would use, but in your case where there is +/- __ meters, you would double the value given to get the range of the uncertainty i...
by Seth_Evasco1L
Sat Oct 28, 2017 5:03 pm
Forum: Formal Charge and Oxidation Numbers
Topic: Formal Charge of Carbon in CH3
Replies: 2
Views: 2792

Formal Charge of Carbon in CH3

In CH3, Carbon has a single bond with each of the three Hydrogen atoms in the molecule, along with one lone electron. What would the formal charge of Carbon be? I'm assuming we don't use that lone electron to represent a lone pair when we use the equation "V-(L+S/2)"?
by Seth_Evasco1L
Thu Oct 26, 2017 10:18 am
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: 2.1
Replies: 4
Views: 1156

Re: 2.1

Jumping from n=1 to n=2 increases the number of shells you have in the atom by 1. Remember that the quantum number "n" represents the principle quantum number/number of shells. For an atom like Helium, it has a principle quantum number of 1 meaning it has one shell which only consists of t...
by Seth_Evasco1L
Thu Oct 26, 2017 10:08 am
Forum: Trends in The Periodic Table
Topic: d5 and d10 e- exception
Replies: 4
Views: 583

Re: d5 and d10 e- exception

All of the elements in the groups containing Copper have the same exception for their electron configuration. However, the only element to have a normal electron configuration in Chromium's group is Tungsten(W).
by Seth_Evasco1L
Thu Oct 26, 2017 10:06 am
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Ground or Excited State?
Replies: 2
Views: 180

Re: Ground or Excited State?

To add on to the previous post: An atom can also be in the excited state if the electrons don't follow Hund's Rule of filling up each empty orbital with electrons of the same spin until all orbitals are filled. For example: (Let's look at Nitrogen) in the 2p subshell where there are 3 orbitals (Px, ...
by Seth_Evasco1L
Sun Oct 22, 2017 11:09 pm
Forum: Quantum Numbers and The H-Atom
Topic: Magnetic Spin Numbers
Replies: 3
Views: 204

Re: Magnetic Spin Numbers

Orbitals can even be filled up initially by all spin-down (-1/2) electrons. The 1/2 and -1/2 electrons both have the same energy so either configuration works as you fill up the orbitals, just as long as the orbitals are filled with electrons that are parallel (same orientation) to each other.
by Seth_Evasco1L
Fri Oct 20, 2017 10:27 am
Forum: Photoelectric Effect
Topic: Energy emitted from an Electron [ENDORSED]
Replies: 4
Views: 326

Re: Energy emitted from an Electron [ENDORSED]

Hi Ishan! The initial energy level should be n=5 because as we've learned, the emission of energy by an electron occurs when that electron relaxes from its excited state down to a lower energy level. So when the electron goes from energy level 5 to energy level 1, energy is emitted and vice versa if...

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