## Search found 56 matches

Wed Mar 14, 2018 11:18 am
Forum: First Order Reactions
Topic: Negative Orders
Replies: 5
Views: 297

### Re: Negative Orders

It can because rate law is experimentally determined so if the experiment yields a result that allows for a negative reaction order then that is possible. Typically I don't think we will be dealing much with negative reaction order because it is not as common but it is good to know that it can exist.
Wed Mar 14, 2018 11:12 am
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: max work & ∆G
Replies: 1
Views: 126

### Re: max work & ∆G

delta G is equal to wmax with a process at constant temperature and pressure I believe. I am not entirely sure.
Wed Mar 14, 2018 11:01 am
Forum: General Rate Laws
Topic: rate law from elementary step
Replies: 2
Views: 148

### Re: rate law from elementary step

I don't think you can use the stoichiometric coefficients from the slow determining step but you can use the reactants and see if what type of molecular reaction it is (i.e. bimolevular, trimolecular, etc.). We use it from the slow-rate determining step because that is the reaction that if it's slow...
Mon Mar 05, 2018 9:58 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: Determining overall rate order
Replies: 3
Views: 179

### Re: Determining overall rate order

Well, rate is usually considered in terms of the consumption of reactants which is easier to measure. So if you had A + B <----> C + D, a hypothetical forward reaction's rate law would like rate = k[A][B], for example. For the reverse reaction, where the reaction mechanism does not change, now it wi...
Mon Mar 05, 2018 9:52 pm
Forum: Zero Order Reactions
Topic: Zero order integrated rate law
Replies: 1
Views: 121

### Re: Zero order integrated rate law

Sure. So here's how it goes: Rate = -d[A]/dt = k[A] 0 for zero order reactions where [A] 0 = 1, so rate = -d[A]/dt = k. Then you have to do a method called separation of variables which is used in calculus where you multiply the dt to the other side to get all the d(stuff) on each side (kind of like...
Mon Mar 05, 2018 9:34 pm
Forum: General Rate Laws
Topic: 15.39b
Replies: 3
Views: 150

### Re: 15.39b

It's because it is specified in the question that the two following equations will be second-order so you just have to use the second-order integrated rate law because it explicitly says that: 1/[A]t = 1/[A]0 + kt.

Manipulate to solve for t and you should get 3.3 x 103 min.

Hope that helps!
Fri Mar 02, 2018 11:30 pm
Forum: Zero Order Reactions
Topic: Relation between rate law and zero order [ENDORSED]
Replies: 3
Views: 163

### Re: Relation between rate law and zero order[ENDORSED]

Yeah for similar problems like that, I suggest looking at the given experimentally determined rate law and then look at all the reactants in the equation and if any reactant is not shown in the rate law shown, then it's most likely zero order because [Reactant]^0 = 1. Hope that helps!
Fri Mar 02, 2018 11:25 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Number of steps to reach the overall reaction
Replies: 3
Views: 176

### Re: Number of steps to reach the overall reaction

Yeah, I don't think it's always the case that the number of steps equals the number of reactants present. I am note sure however.
Fri Mar 02, 2018 11:23 pm
Forum: General Rate Laws
Topic: Integrated Rate Laws when a =/= 1
Replies: 4
Views: 212

### Re: Integrated Rate Laws when a =/= 1

I am not sure if I will be able to exactly answer your question, but I think each order has it's own integrated rate law. To find the order, I suggest using the given concentrations and if time is also given, then plot each type of integrated rate law and see what the order is based off of that.
Fri Mar 02, 2018 11:23 pm
Forum: General Rate Laws
Topic: Integrated Rate Laws when a =/= 1
Replies: 4
Views: 212

### Re: Integrated Rate Laws when a =/= 1

I am not sure if I will be able to exactly answer your question, but I think each order has it's own integrated rate law. To find the order, I suggest using the given concentrations and if time is also given, then plot each type of integrated rate law and see what the order is based off of that.
Sat Feb 24, 2018 1:24 pm
Forum: General Rate Laws
Topic: Stoichiometric Coefficient [ENDORSED]
Replies: 2
Views: 146

### Re: Stoichiometric Coefficient[ENDORSED]

I am not sure but based on what we have seen in class, then yes.
Sat Feb 24, 2018 1:23 pm
Forum: Balancing Redox Reactions
Topic: 14.25
Replies: 3
Views: 166

### Re: 14.25

I don't have the textbook in front of me but if you could post the question, it might be easier. Thanks!
Sat Feb 24, 2018 1:22 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 14.15c [ENDORSED]
Replies: 3
Views: 157

### Re: 14.15c[ENDORSED]

Yeah, I am confused on that too. I don't remember the exact question, but I think if it's in a basic solution you might choose something like KOH to act as a salt bridge? Hopefully someone else can explain better than me.
Thu Feb 15, 2018 10:56 pm
Forum: Student Social/Study Group
Topic: Post All Chemistry Jokes Here
Replies: 7774
Views: 1073233

### Re: Post All Chemistry Jokes Here

I heard O-chem is difficult. Those who study it have alkynes of trouble.
Thu Feb 15, 2018 10:48 pm
Forum: Student Social/Study Group
Topic: Post All Chemistry Jokes Here
Replies: 7774
Views: 1073233

### Re: Post All Chemistry Jokes Here

Disclaimer: A Corny Acid-Base joke
What did H2PO4- say to H2O?

"Why you gotta be so BiPoLaAr!"

Water: "As an amphoteric molecule, I like to have multiple affairs!"

H2PO4-: "Can't believe I am dating a weirdo."
Thu Feb 15, 2018 10:35 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Homework #8.55 Hess's Law
Replies: 2
Views: 127

### Re: Homework #8.55 Hess's Law

I am not entirely sure but whichever one has the correct balanced equation (I think it's the one in solutions manual?). It makes sense for 3/2 to be a coefficient for O2 to get 3 moles of oxygen in the product. Hope that helps a little!
Wed Feb 07, 2018 10:50 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: Molar Entropy
Replies: 2
Views: 98

### Re: Molar Entropy

Yes. Technically speaking, the bigger the molecule is implies that the molar entropy will be higher because there are more atoms that can have different positional arrangements (microstates - W), more atoms means more bond vibration energy, and a higher rotational energy etc. all contributing to the...
Wed Feb 07, 2018 11:47 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: unit of standard reduction potential
Replies: 2
Views: 94

### Re: unit of standard reduction potential

Voltage is the unit! Remember to help with conversions, 1 J/C (C=coulumb) = 1V.
Wed Feb 07, 2018 11:42 am
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: 9.45
Replies: 3
Views: 171

### Re: 9.45

Yes, but you use -delta Hfus for delta H because it's freezing, not melting and keep the temperature the same. Hope that helps!
Sat Feb 03, 2018 5:39 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: ΔS(universe)
Replies: 6
Views: 270

### Re: ΔS(universe)

Hi, So delta S according to the Second Law of Thermodynamics will always be increasing in the universe. Delta S of the system can decrease, in other words, if you freeze liquid water to ice, you are reducing the possible number of arrangements (microstates) that the H 2 0 molecule can be in. So delt...
Sat Feb 03, 2018 5:33 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Value of G
Replies: 4
Views: 129

### Re: Value of G

I am not sure exactly on the lowest value of G, but I do know that delta G can have a positive value, negative value, and a value of 0 because it is a state function so you have to think about final minus initial.
Sat Feb 03, 2018 5:21 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 11.15 Gibbs Free Energy and Q
Replies: 1
Views: 101

### Re: 11.15 Gibbs Free Energy and Q

Hi,

Yeah so to find Q, you can just plug in the partial pressures which is directly proportional to concentrations, so Q = (.98)^2/(.13). I hope that helps! That should then give you the right answer hopefully.
Tue Jan 23, 2018 2:11 am
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Gibbs free energy equations
Replies: 4
Views: 201

### Re: Gibbs free energy equations

I think they are essentially the same when we do the computation. The delta G of formation is the energy it takes to make the molecule you want to find from its base elements, which is different from molar Gibbs' free energy which is for molecules in their standard states at 1 mole (it's not necessa...
Tue Jan 23, 2018 2:06 am
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: q isn't a state function
Replies: 3
Views: 233

### Re: q isn't a state function

Heat results because of a temperature difference. We know temperature is a state function because you can do T (final) - T (initial), but it doesn't really make sense to find delta q = q(final) - q(initial) because q is dependent on the pathway. Think of it like this: Every molecule motion due to th...
Tue Jan 23, 2018 2:01 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Work Done On or By System
Replies: 1
Views: 131

### Re: Work Done On or By System

Work done by system: negative sign for work Work done on system: positive sign for work. Another way to help differentiate the two: Think of the equation: w= -P * (delta V). ---->If volume expansion occurs, then final volume minus initial volume is going to be positive so with the negative sign in t...
Mon Jan 15, 2018 2:17 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: 8.25
Replies: 1
Views: 135

### Re: 8.25

-q cal = q (of the main reaction). You can find delta U from there since w=0 because there's constant volume (delta V is 0) so delta U is equal to q. There's a thread on this same question too which I answered in a little bit more detail and can check out here: https://lavelle.chem.ucla.edu/forum/vi...
Mon Jan 15, 2018 2:13 pm
Forum: Calculating Work of Expansion
Topic: Question 8.11 Reversible v Irreversible actions
Replies: 2
Views: 173

### Re: Question 8.11 Reversible v Irreversible actions

In part a, the process is irreversible meaning there is an external, constant pressure and the amount of work is usually less than in part b, in which it's a reversible process, where more work is done. Equation for part a to use: w = -P*deltaV and for part b it's: w = -nRT ln (V f /V i ). In a reve...
Mon Jan 15, 2018 2:03 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Ch. 8 #15 Part d
Replies: 2
Views: 178

### Re: Ch. 8 #15 Part d

The way I understood was that there can be heat present in the system already but no further heat transfer is allowed; only work energy is transferable. Since delta U = q + w, then U = q if w = 0.

Hope that helps!

Sincerely,
Yashwi
Mon Jan 08, 2018 9:16 pm
Forum: Phase Changes & Related Calculations
Topic: Question on 8.21
Replies: 3
Views: 151

### Question on 8.21

Here's Question 8.21: A piece of copper of mass 20.0 g at 100.0 degrees Celsius is placed in a vessel of negligible heat capacity but containing 50.7 g of water at 22.0 degrees Celsius. Calculate the final temperature of the water. Assume that no energy is lost to the surroundings. I understand how...
Mon Jan 08, 2018 8:53 pm
Forum: Phase Changes & Related Calculations
Topic: Question 8.25
Replies: 4
Views: 237

### Re: Question 8.25

I think the value that we really need to find here is change in internal energy (delta U). We know that delta U is equal to q + w. Since the question involves a bomb-calorimeter (constant volume), we know that w should equal zero since w=-P*deltaV and delta V is zero. So we only need to really find ...
Mon Jan 08, 2018 8:35 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: 8.3 Reversible Isothermal Expansion
Replies: 3
Views: 240

### Re: 8.3 Reversible Isothermal Expansion

Hi, I think you brought up a good point. I am a bit confused by your question but I think what they are saying essentially is during a reversible process, a system is in thermodynamic equilibrium with its surroundings. So for every reduction in external pressure, the volume usually changes infinites...
Mon Dec 04, 2017 3:11 am
Forum: Lewis Acids & Bases
Topic: Donating electrons versus protons [ENDORSED]
Replies: 4
Views: 257

### Re: Donating electrons versus protons[ENDORSED]

I think an easier way to think of it is like this: If a molecule has electron lone pairs on it, according to the Lewis definition, it will be a Lewis base because it is donating an electron pair, while the Lewis Acid is the molecule that accepts the electron lone pair. So for instance, if I have: NH...
Mon Dec 04, 2017 2:58 am
Forum: Properties & Structures of Inorganic & Organic Acids
Topic: J5. Complete Ionic Equations
Replies: 1
Views: 127

### Re: J5. Complete Ionic Equations

Hi, So generally weak acids or weak bases when they are mixed with water, they don't dissociate all their protons and that's why we write them as molecules instead of separating them like we do for Ca(OH)2 for example. HF is a weak acid so we leave it, we don't need to separate it. Hope that helps! ...
Mon Nov 27, 2017 10:03 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Terminology [ENDORSED]
Replies: 1
Views: 145

### Re: Terminology[ENDORSED]

Using shift is fine, but it is better to describe the reaction as either sitting to the left, if there are more reactants, or sitting to the right, if there are more products. Maybe, I am not sure considering we didn't do LeChatelier's Principle yet, if you have to describe which way the reaction wi...
Mon Nov 27, 2017 9:58 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: HW Question 11.43 and general question
Replies: 1
Views: 118

### Re: HW Question 11.43 and general question

Hi, I don't have the textbook with me right now. Maybe if you could please post the question up? For your general question, say I have 3H 2 + N 2 <---> 2NH 3 . To set up the ice table, let's assume that we have 0 NH 3 initially and want to form that product. In the change section of the ICE table, w...
Mon Nov 20, 2017 9:57 am
Forum: Shape, Structure, Coordination Number, Ligands
Topic: Chelating Complexes
Replies: 5
Views: 263

### Re: Chelating Complexes

@Michael So a chelate is, by definition from the textbook: "A complex containing one or more ligands that form a ring of atoms that includes the central metal atom. " Basically a chelate is like a name for a type of structure where a ring of ligands are attached to the central metal atom. ...
Mon Nov 20, 2017 9:43 am
Forum: Naming
Topic: 17.31d
Replies: 4
Views: 297

### Re: 17.31d

Hi yes, So these are the steps that I like to follow: 1) Identify the cation and the anion and make sure you write cation first then the anion. In this case Sodium is the cation and the anion is bisoxalato(diaqua)ferrate (III) anion. 2) Focus on writing the anion first and make sure you open bracket...
Fri Nov 17, 2017 3:23 pm
Forum: Hybridization
Topic: axial vs equatorial lone pair
Replies: 2
Views: 257

### Re: axial vs equatorial lone pair

Yes, If you are taking out an electron from a trigonal bipyramidal shape, then take it from the equatorial position so that the lone-pair electron can have room for itself and push the other equatorial bonds closer to each other. If it is however, an octahedral shape, then remember that there are 6 ...
Fri Nov 17, 2017 3:16 pm
Forum: Hybridization
Topic: 4.35
Replies: 2
Views: 140

### Re: 4.35

I believe they are a similar question. If it asks to state/identify, you can provide sp^3 for CH4 because it has 4 electron regions as a tetrahedral figure. Maybe the solutions manual could help if you are still stuck on it. Hope that helps!

-Yashwi
Fri Nov 17, 2017 3:12 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: 4.33
Replies: 2
Views: 163

### Re: 4.33

It is not required to write out the coefficient I believe, but it would be better practice to do so. As long as the hybridization matches up with the number of electron regions. For instance sp^3 = for 4 electron region densities.
Fri Nov 10, 2017 2:20 pm
Forum: Lewis Structures
Topic: Multiple lewis structures for radicals [ENDORSED]
Replies: 3
Views: 219

### Re: Multiple lewis structures for radicals[ENDORSED]

An easy way to help figure out where the lone electron goes is to see which element will have a formal charge close to zero depending on the placement of the lone electron. For instance, in CH3, the lone electron would go on the C since the formal charge would be 0 as it is more favorable too. H can...
Wed Nov 08, 2017 9:30 pm
Forum: Dipole Moments
Topic: Direction of the Dipole moment arrow [ENDORSED]
Replies: 3
Views: 247

### Direction of the Dipole moment arrow[ENDORSED]

Hi, I know in lecture we mentioned that the arrow for the dipole moment points toward the negative partial charge with the plus sign on the positive partial charge. In the textbook, it mentions that the way I described above is the conventional method and the book will use the modern convention with...
Tue Oct 31, 2017 10:03 pm
Forum: Trends in The Periodic Table
Topic: Writing Electron Config for Chromium after being ionized once?
Replies: 2
Views: 197

### Writing Electron Config for Chromium after being ionized once?

Hi, so I know that chromium's e-config is: [Ar](3d^5)(4s^1) as it is one of the exceptions to the e-configuration. When that element is ionized and becomes an ion (Cr^+1), which is the correct e-config?: (1) [Ar](3d^5)...because the electron is removed from the s-orbital? or (2) [Ar](3d^4)(4s^1)...b...
Tue Oct 31, 2017 9:54 pm
Forum: Trends in The Periodic Table
Topic: Ionization energy exception; O
Replies: 2
Views: 837

### Re: Ionization energy exception; O

Basically, atoms love to either have completely filled orbitals or half-filled orbitals. Nitrogen loves its half-shells, so it will be very hard to remove an electron, thus it will take a high amount of energy to remove one of the electrons from a half-orbital, making Nitrogen have a high ionization...
Wed Oct 25, 2017 10:19 pm
Forum: Trends in The Periodic Table
Topic: Electron Affinity and Ionization Energy
Replies: 3
Views: 321

### Re: Electron Affinity and Ionization Energy

Yes, electron affinity is the energy that's released when an electron is added. Ionization energy is energy needed to remove an electron. Really important to know (maybe not for now but helpful for later too) is the following: Electron Affinity (Mostly Exothermic)...For the first electron affinity e...
Wed Oct 25, 2017 10:18 pm
Forum: Trends in The Periodic Table
Topic: Electron Affinity and Ionization Energy
Replies: 3
Views: 321

### Re: Electron Affinity and Ionization Energy

Yes, electron affinity is the energy that's released when an electron is added. Ionization energy is energy needed to remove an electron. Really important to know (maybe not for now but helpful for later too) is the following: Electron Affinity (Mostly Exothermic)...For the first electron affinity e...
Wed Oct 25, 2017 10:03 pm
Forum: Trends in The Periodic Table
Topic: f-orbitals
Replies: 6
Views: 1612

### Re: f-orbitals

The f-orbitals start with the lanthanides and the actinides, in the n=6 row and the f-orbitals have an n=4 as the principle quantum energy number. Specifically, the f-orbitals start with the element 58. n=4 is the energy level, b/c I believe it's in a lower energy state than the d-orbitals, which is...
Wed Oct 25, 2017 11:24 am
Forum: Electron Configurations for Multi-Electron Atoms
Topic: HW Question 2.37
Replies: 2
Views: 126

### Re: HW Question 2.37

I am not exactly sure but I think for an electron to penetrate a nucleus means the electron can be attracted to the positive force of the nucleus and in effect shields the outermost electrons. Hope this helps!
Wed Oct 25, 2017 11:08 am
Forum: Electron Configurations for Multi-Electron Atoms
Topic: What exactly is an excited state? [ENDORSED]
Replies: 3
Views: 256

### What exactly is an excited state?[ENDORSED]

Hi, I know the ground state configurations for elements but does the excited state mean any deviation from the ground-state configuration? My interpretation of an excited state was that some electrons will jump to higher subshells or orbitals. In the textbook in one of the problems, they showed the ...
Fri Oct 20, 2017 11:40 am
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Spin of parallel electrons
Replies: 1
Views: 178

### Spin of parallel electrons

In class today, we discussed how the spin of parallel electrons can either be up or down. Whenever I draw electron configurations for ex. Nitrogen's, for the p subshell, I draw the electrons as if their m(s) is +1/2. For ex: for nitrogen I would do: [He] (in 2s-orbital: up arrow and down arrow) (in ...
Fri Oct 20, 2017 11:21 am
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: Formal Charge
Replies: 3
Views: 403

### Re: Formal Charge

When looking at a lewis structure diagram, there will be lone pair electrons (dots around the element) which you will count each dot as one and then there will also be the single bond lines, which instead of counting as two electrons, count them as one. So add them up and then subtract it from the v...
Fri Oct 13, 2017 10:10 pm
Forum: DeBroglie Equation
Topic: Post-Module Assessment Q. 34
Replies: 4
Views: 268

### Re: Post-Module Assessment Q. 34

I am not exactly sure but as far as I know, the speed of light is the fastest speed on earth, unless research shows otherwise. Thus, if you get a speed less than 3.00 * 10^8 m/s, I am pretty sure it's reasonable b/c it's less than speed of light, which is the fastest speed known so far. That's my th...
Fri Oct 13, 2017 11:05 am
Forum: Einstein Equation
Topic: General Question about Joules [ENDORSED]
Replies: 7
Views: 663

### Re: General Question about Joules[ENDORSED]

Hi,

A Joule is the definition as mentioned above. However, to help with your calculations and unit conversions, A Joules is also equal to kg((m/s)^2). Hope that helps!
Tue Oct 10, 2017 10:16 pm
Forum: DeBroglie Equation
Topic: Post-Module Assessment Q. 34
Replies: 4
Views: 268

### Re: Post-Module Assessment Q. 34

I think your answer is like correct, meaning approx. close to the answers given b/c when you round the answer you got and change it to scientific notation, you should get 1.00 * 10^5 m/s and since it's way less than speed of light, which if the fastest speed known so far, answer should be reasonable...
Thu Oct 05, 2017 10:53 pm
Forum: Balancing Chemical Reactions
Topic: Writing chemical formulas [ENDORSED]
Replies: 5
Views: 430

### Re: Writing chemical formulas[ENDORSED]

Hi, Just a quick question...when writing chemical reactions and balancing them, if there is a catalyst involved or heat being utilized for the reaction to occur, do we have to also place the catalyst's name above the arrow and place the triangle (representing heat being involved) above the arrow too...
Thu Oct 05, 2017 1:42 pm
Forum: Limiting Reactant Calculations
Topic: Combustion analysis help?
Replies: 3
Views: 285

### Re: Combustion analysis help?

Sure. So Combustion Reactions generally involve organic compounds or hydrocarbons (they can also involved other molecules too) that burn in the presence of oxygen. So whenever you hear the words burning, combustion, etc. always think of the compound given reacting with oxygen gas (O2). Now, the prod...
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