Search found 29 matches

Mon Mar 07, 2016 9:58 am
Forum: General Rate Laws
Topic: question 15.23
Replies: 1
Views: 324

Re: question 15.23

Its the exact same thing. What they did was that they moved the negative over to the left hand side and used log rules to move that -1 into the exponent of what is inside the reaction. Therefore, you get something like ln[ (Afinal/Ainitial)^-1]. The -1 as an exponent flips the two of them and then y...
Wed Mar 02, 2016 2:37 pm
Forum: *Organic Reaction Mechanisms in General
Topic: Kinetic stability
Replies: 1
Views: 367

Re: Kinetic stability

You can see the relationship by looking on a graph with delta G on the y axis and time on the x axis (free energy diagram). The delta G for the transition state is what we are focused on so that delta G can be found by looking at the distance between the reactants and the peak of the curve. A small ...
Mon Feb 22, 2016 12:24 pm
Forum: *Free Energy of Activation vs Activation Energy
Topic: Activation Energy with Relation to k
Replies: 1
Views: 389

Re: Activation Energy with Relation to k

Conceptually, if the activation energy is high, then the reactants would need to have a large amount of energy in order to overcome the activation energy and start forming the products. Less reactants are likely to have a large amount of energy necessary to overcome the activation energy and therefo...
Thu Feb 18, 2016 9:50 pm
Forum: Kinetics vs. Thermodynamics Controlling a Reaction
Topic: Calculating the value of the rate constant
Replies: 1
Views: 433

Re: Calculating the value of the rate constant

You would just use the Arrhenius Equation but twice. Basically what you want to do is have \frac{k(temp1)}{k(temp2)}=\frac{e^{\frac{Ea}{RT}}}{e^\frac{Ea}{RT}} Where Ea is the same for both k (activation energy remains constant at different temperatures) and R is the gas constant. Als...
Mon Feb 15, 2016 9:34 am
Forum: General Rate Laws
Topic: Units of k
Replies: 2
Views: 331

Re: Units of k

It is because the rate law is third order (second order with respect to A and first order with respect to C). Therefore, you end up with units \frac{mol^3}{L^3} . Since this is a rate law, you want the units to become \frac{mol}{Ls} . In order to do so, you need to get rid of some of those units by ...
Tue Feb 09, 2016 8:15 pm
Forum: Phase Changes & Related Calculations
Topic: 2015 Practice Midterm Q1A
Replies: 1
Views: 345

Re: 2015 Practice Midterm Q1A

Since it does not state that the water turns into steam in the question, it means that it is just asking for the amount of energy needed to bring the water up to 100C and no more. Water doesn't start to boil because it is at 100C. It only boils when more energy is added to the water at 100C. If you ...
Mon Feb 01, 2016 9:08 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: inert electrodes
Replies: 1
Views: 289

Re: inert electrodes

You include one if one part of the redox reaction does not have a conductive solid counterpart. For example, MnO4- is an ion and you can't have a piece of pure MnO4- solid because it is an aqueous solution. Therefore, you would need an inert electrode like platinum to carry the charge.
Mon Jan 25, 2016 9:57 am
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Residual entropy
Replies: 1
Views: 459

Re: Residual entropy

Residual entropy is the entropy at which temperature = 0K for the crystalline form of the compound. You can calculate the residual entropy of a compound using the Boltzmann equation S = kb*lnW. All you have to do is determine the degeneracy of the crystalline version of the molecule. This means that...
Tue Jan 19, 2016 10:38 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 8.57 Enthalpy Units
Replies: 1
Views: 576

Re: 8.57 Enthalpy Units

It is because the enthalpy is for "one reaction". Basically, the enthalpy you calculated is the enthalpy for that specific reaction as it is written. So if you were given 2C2H2 + 4H2 -> 2C2H6 you would find that the enthalpy of that written reaction is two times the enthalpy of the reactio...
Mon Jan 11, 2016 9:32 am
Forum: Calculating Work of Expansion
Topic: Internal energy change
Replies: 1
Views: 301

Re: Internal energy change

Yes because the internal energy is the potential/kinetic energy of the molecules. Basically it deals with the motion of molecules. Temperature is the average kinetic energy of the atoms in the object. So while the two are not directly proportional with each other because internal energy includes pot...
Fri Jan 08, 2016 11:13 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Standard Enthalphy of a reaction
Replies: 3
Views: 401

Re: Standard Enthalphy of a reaction

In lecture, he said that if the compounds were not in standard state then we were suppose to treat it as if the compounds were in standard state and then adjust for the phase changes. I'm not sure how to do this but I think it involves taking difference between the amount of energy in the compounds ...
Fri Dec 04, 2015 10:54 pm
Forum: Acidity & Basicity Constants and The Conjugate Seesaw
Topic: Stronger acid in acetic acid?
Replies: 1
Views: 747

Re: Stronger acid in acetic acid?

This is all about Le Chatlier's Principle. I'm not sure if the following chemical equation is right but: HNO2 + CH3COOH <-> 2H+ + NO2- + CH3COO- A strong base is something that easily takes up a proton. Because NO2 is a weaker base then CH3COO-, that means that CH3COO- is more likely to take up the ...
Wed Dec 02, 2015 11:05 pm
Forum: *Titrations & Titration Calculations
Topic: Calculating titrations beyond stoichiometric point
Replies: 3
Views: 863

Re: Calculating titrations beyond stoichiometric point

Because at the stoichiometric point, all of the base has been neutralized by the acid and leaving behind only a salt and water. Since NaOH is a strong base and HCl are strong acids, the salts that are formed do not affect the pH. When you keep adding more acid, all you're doing is adding more H30+ s...
Tue Dec 01, 2015 12:14 am
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Favoring products or reactants
Replies: 1
Views: 341

Re: Favoring products or reactants

The course reader describes situations where the products or reactants will be "strongly favored". Quiz 3 simply asked if either the products or reactants are favored, not necessarily strongly favored. Since the K > 1, it can be said that the concentration of products are greater than the ...
Thu Nov 26, 2015 9:08 pm
Forum: Bronsted Acids & Bases
Topic: Concentration of ions in a solution?
Replies: 1
Views: 351

Re: Concentration of ions in a solution?

HI is a strong acid so we can assume that there will be 100% dissociation. Therefore, a concentration of 6x10^-5 M HI would yield 6x10^-5 M of H+ (H30+). And then you can use the formula pH + pOH = 14 to get the concentration of OH-.
Mon Nov 23, 2015 7:55 am
Forum: Acidity & Basicity Constants and The Conjugate Seesaw
Topic: pKa and pKb and more
Replies: 1
Views: 441

Re: pKa and pKb and more

The p stands for $-log{_{10}}$
So pKa is $-log_{10}K_{a}$
Sat Nov 21, 2015 9:23 pm
Forum: Lewis Acids & Bases
Topic: Determining Strong Acids and Bases
Replies: 2
Views: 491

Re: Determining Strong Acids and Bases

I think that we would probably have to know some of the basic strong acids and strong bases. There is a small list of them on pg 151 of the course reader. I don't know any method of memorizing the strong acids except to just memorize them but for the bases you can imagine like a lowercase letter &qu...
Wed Nov 18, 2015 10:52 pm
Forum: Bronsted Acids & Bases
Topic: Acid Strength?
Replies: 2
Views: 398

Re: Acid Strength?

I think that the electronegativity of F would pull that lone electron from the H towards itself and the H itself would be pulled closer to the F making it a shorter bond and harder to dissociate (not quite sure if this is how it works). But the main thing is that the atomic radius of F is so much sm...
Sat Nov 14, 2015 7:58 pm
Forum: Ideal Gases
Topic: Real Gases Clarification
Replies: 1
Views: 396

Re: Real Gases Clarification

The first part is right. As pressure goes up volume decreases. The most ideal gas will be the one with the lowest pressure and the highest temperature. That is because in an ideal gas, the molecules do not exert any forces between each other. That is, the intermolecular force is negligible. This is ...
Thu Nov 12, 2015 11:20 pm
Forum: Ideal Gases
Topic: Using R to find concentration
Replies: 3
Views: 594

Re: Using R to find concentration

PV=nRT when manipulated equals P=nRT/V
Concentration = number of moles/volume = n/V You substitute this value into the equation and you get:
P=MRT
This should allow you to solve for the concentration in the problem.
Wed Nov 11, 2015 10:45 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Conditions favoring production - pressure
Replies: 6
Views: 813

Re: Conditions favoring production - pressure

An increase in pressure would result in favoring the side with fewer moles while a decrease in pressure would result in favoring the side with more moles. This is assuming that the change in pressure is done by changing the volume of the system and not by adding inert gases.
Tue Nov 03, 2015 3:16 pm
Forum: Ionic & Covalent Bonds
Topic: Why is LiH more ionic than HCl?
Replies: 2
Views: 743

Re: Why is LiH more ionic than HCl?

For electronegativity trends, the position of Hydrogen is incorrect on the periodic table. It turns out hydrogen has an electronegativity of 2.1ish and really should be placed on top of Boron on the periodic table. When that is the case, Hydrogen is closer to Chlorine then to Lithium. eneg6.jpg Hope...
Fri Oct 30, 2015 10:26 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: How to tell unpaired electrons from electron configurations
Replies: 1
Views: 1375

Re: How to tell unpaired electrons from electron configurati

It's a little hard to tell determine how many unpaired electrons based off just the electron configuration. It's easier if you draw or imagine the aufbau diagram like (1s _ 2s_ 2p_ _ _). For example Carbon has an aufbau diagram of 2p _ _ _ 2s_ 1s_ And then you just fill in electrons. Carbon has six ...
Sat Oct 24, 2015 8:47 pm
Forum: Ionic & Covalent Bonds
Topic: ground state electron configurations
Replies: 3
Views: 1692

Re: ground state electron configurations

You take off the electrons that are the farthest away from the nucleus first. The p orbital electrons are farther away from the nucleus then the s orbital electrons and in the example it is the farthest away: (Ground state of Sb is [Kr]4d10 5s2 5p3) Sb3+ would mean taking three electrons away so tha...
Fri Oct 16, 2015 8:36 pm
Forum: Lewis Structures
Topic: Lewis Structures - Lone Pairs
Replies: 1
Views: 265

Re: Lewis Structures - Lone Pairs

Where lone pairs are located are arbitrary because we don't know exactly where that electron is in the orbitals although they should just be on the nitrogen. And it's the other way around. We take lone pairs into account when we figure out what the molecular shapes are. In this case, there are only ...
Fri Oct 16, 2015 8:23 pm
Forum: Resonance Structures
Topic: Angstrom Calculation
Replies: 2
Views: 752

Re: Angstrom Calculation

Are you talking about how they got the number 1.24 ‎Å?
If you are, then it is just found experimentally through x-ray diffraction.
Fri Oct 09, 2015 9:09 pm
Forum: Properties of Light
Topic: Wavelength
Replies: 1
Views: 398

Re: Wavelength

The formula you use is similar to that one. The formula is \upsilon = R(\frac{1}{n_{1} ^2}-\frac{1}{n_{2} ^2}) The first n is 3 and the second n is 4. The reason is because a photon is released when an electron falls from a higher energy level to a lower energy level. The equation is based o...
Wed Oct 07, 2015 10:05 am
Forum: Quantum Numbers and The H-Atom
Topic: Chapter 2, Problem 19, part A # of values for l
Replies: 1
Views: 380

Re: Chapter 2, Problem 19, part A

The quantum number, l, describes the shape of the orbital and is called the subshell. For example, 0 would be spherical. The shapes get more and more complex as l continues to increase. There are more descriptions of orbitals then s, p, d, f. It can be g, h...etc. It's just that the we are unlikely ...
Thu Oct 01, 2015 11:28 am
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: How many photos of infrared radiation?
Replies: 1
Views: 330

Re: How many photos of infrared radiation?

It sounds like you're doing everything right. The 60W is extraneous because they are only talking about infrared radiation which is only a part of the types of electromagnetic waves a lightbulb can emit (like visible light).
Hope this helps!

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