## Search found 20 matches

Fri Mar 17, 2017 3:23 pm
Forum: *Cyclohexanes (Chair, Boat, Geometric Isomers)
Topic: Stability
Replies: 1
Views: 222

### Re: Stability

The most stable conformation for a cyclohexane is the chair conformations with your largest substituent bonding to the highest carbon through an equatorial bond. There is also the boat conformation that is not stable at all because of the steric interactions.
Mon Mar 06, 2017 12:30 pm
Forum: *Alkanes
Topic: Molecular Formula and Chemical Structure
Replies: 2
Views: 317

### Re: Molecular Formula and Chemical Structure

When dealing with molecules that have four or less carbons the chemical structure can only be unbranched in order to accommodate all carbons and hydrogens in the molecule. When a molecule contains 5 or more carbons the molecule's structure can begin to take various forms. Each of these forms are equ...
Thu Mar 02, 2017 12:28 pm
Forum: *Free Energy of Activation vs Activation Energy
Topic: Delta G Double Dagger, Delta H Double Dagger, Delta S Double Dagger
Replies: 3
Views: 2732

### Re: Delta G Double Dagger, Delta H Double Dagger, Delta S Double Dagger

I'm pretty sure that the delta G and delta H double daggers are always going to be positive because they represent the amount of energy that is needed to overcome the activation barrier. If you have a reaction A+B->AB(double dagger)->C energy is required to first break the bonds of A and B so that t...
Tue Feb 21, 2017 1:34 pm
Forum: Kinetics vs. Thermodynamics Controlling a Reaction
Topic: Homework 15.79
Replies: 4
Views: 426

### Re: Homework 15.79

Will we need to know these types of problems for Quiz 2 or can we disregard any problem that has to do with temperature?
Fri Feb 17, 2017 2:54 am
Forum: Balancing Redox Reactions
Topic: Balancing (Acidic vs. Basic Balancing)
Replies: 3
Views: 442

### Re: Balancing (Acidic vs. Basic Balancing)

The only difference between balancing an acidic and basic solution is that the basic solution can't contain any H+1 so you would just multiply equation through by an OH-1.
Wed Feb 08, 2017 5:42 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: Understanding n intuitively
Replies: 1
Views: 182

### Re: Understanding n intuitively

Just compare the concentration of one of the reactants, say [NH4+1] to the rate. After you've recognized that as the concentration doubles your rate doubles you can say that 2^n=2. Therefore, n must equal 1. This logic provides you with a shortcut.Hope this helps.
Mon Jan 30, 2017 2:30 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Problem 8.31
Replies: 1
Views: 207

### Re: Problem 8.31

The units for delta T are arbitrary because the difference between your final T and initial T are going to be the same whether you are calculating it in celsius or kelvin.
Tue Jan 24, 2017 12:47 pm
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: Why are W and V interchangeable in the Boltzman formula?
Replies: 2
Views: 222

### Re: Why are W and V interchangeable in the Boltzman formula?

I believe because W is simply the number of different ways or combinations in which atoms can in a sense exist. Using that definition of W we can assume that if the volume has doubled then so have the number of different ways in which the atoms can orient themselves. If V2/V1 = 2/1 then W2/W1 = 2/1.
Wed Jan 18, 2017 5:27 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: ch.8/#25
Replies: 2
Views: 310

### Re: ch.8/#25

I am pretty sure that q of the reaction + q of the calorimeter = 0 because the heat given off by the reaction is the same heat that is being absorbed by the calorimeter.
Sat Jan 14, 2017 4:16 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Chapter 8 #59 [ENDORSED]
Replies: 3
Views: 434

### Re: Chapter 8 #59[ENDORSED]

The enthalpy of N2 is 0 because N2 is nitrogen in its most stable form (nitrogen is found as N2 in nature).
Fri Dec 02, 2016 2:16 pm
Forum: Photoelectric Effect
Topic: Midterm 3B
Replies: 1
Views: 319

### Re: Midterm 3B

That the amount of photons is at times irrelevant. You can have an innumerable amount of photons but they wouldn't excite an electron unless each photon has at least a specific amount of energy. The photons are being quantized.
Sat Nov 26, 2016 9:45 am
Forum: Acidity & Basicity Constants and The Conjugate Seesaw
Topic: Number 33 in the Chapter 12 HW
Replies: 1
Views: 261

### Re: Number 33 in the Chapter 12 HW

A) -log[OH-1]=13.25 and solve for [OH-1]
B)once you get [OH-1] use M1V1=M2V2 to find concentration of OH-1 in original solution.
C) (0.2liters/1)(moles of OH-1 which is just molarity value/1 liter)(molar mass of Na2O/1 mol Na2O)
Tue Nov 15, 2016 9:49 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: K vs Q [ENDORSED]
Replies: 4
Views: 493

### Re: K vs Q[ENDORSED]

K is simply the value calculated when the reaction is at equilibrium. You use the concentrations of the reactants and products at equilibrium. Q is pretty much any other point in time for the reaction.
Tue Nov 08, 2016 4:21 pm
Forum: *Molecular Orbital Theory (Bond Order, Diamagnetism, Paramagnetism)
Topic: MO Correlation Diagrams for Heteronuclear molecules
Replies: 1
Views: 269

### Re: MO Correlation Diagrams for Heteronuclear molecules

Like you said the more electronegative atom's orbitals is drawn slightly lower than the less electronegative atom's orbitals. When you draw the lines that represent your 2s or 2p orbitals make sure they are lower for the more eneg atom. It should look a bit asymmetric.
Thu Nov 03, 2016 11:52 am
Forum: Properties of Light
Topic: What to do when a Energy is given in kJ*mol-1?
Replies: 6
Views: 807

### Re: What to do when a Energy is given in kJ*mol-1?

That is exactly what you do. You're given the energy per mol but you want the total energy which is why you would do that conversion.
Thu Oct 27, 2016 11:04 am
Forum: Formal Charge and Oxidation Numbers
Topic: Formal Charge for Indivual Elements in Compound
Replies: 2
Views: 621

### Re: Formal Charge for Indivual Elements in Compound

Normally the more electronegative atom would have the negative charge. This makes sense because they are holding the shared electron closer to their nucleus than the other atom is. Electron has a negative 1 charge which contributes to a negative formal charge for the more electronegative atom.
Wed Oct 19, 2016 10:57 am
Forum: Trends in The Periodic Table
Topic: Electron affinity
Replies: 5
Views: 579

### Re: Electron affinity

Br usually has a -1 charge because it has 7 valence electrons. Electrons want to have a full valence shell so Bromine will usually "steal" an electron from an atom with less electron affinity("weaker"). With that being said if Bromine bonds with nonmetal higher up on the table it...
Wed Oct 12, 2016 2:53 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Clarification on ENERGY LEVEL concept
Replies: 3
Views: 1778

### Re: Clarification on ENERGY LEVEL concept

When looking for the quantum level it's important to realize that if the level is decreasing then you are going to have a negative energy change. That's why when you use the change in energy = -hr/n^2final - (- hr/n^2initial) you have to make sure your change in energy is consistent with whether n i...
Thu Oct 06, 2016 12:37 pm
Forum: Empirical & Molecular Formulas
Topic: Fundamentals M.25 Help [ENDORSED]
Replies: 1
Views: 349

### Re: Fundamentals M.25 Help[ENDORSED]

So you know that if the only molecule present was the 2-naphthol your percent of carbon would be 10(because you have 10 carbon atoms) times 12 grams/mol of carbon all divided by 144.16 g/mol( molar mass of 2-naphthol). Multiply whatever decimal you get by 100 and you well get your percentage of carb...
Mon Sep 26, 2016 2:40 pm
Forum: Molarity, Solutions, Dilutions
Topic: Fundamentals G5
Replies: 6
Views: 1101

### Re: Fundamentals G5

So the molarity of Na2CO3 is 0.08 M according to your calculations. Because for every one mole of Na2CO3 you have 2 moles of Na+1 multiply 0.08 M by 2 and that is the molarity of Na+1. Take 0.00215 mol/ 0.08 M times 2 and that should give you your answer. Makes sense too because units cancel out to ...
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