## Search found 52 matches

- Tue Mar 13, 2018 12:33 pm
- Forum: *Enzyme Kinetics
- Topic: Catalysis 15.69
- Replies:
**3** - Views:
**214**

### Re: Catalysis 15.69

You can set up a proportion of k of the catalyzed reaction and k of the uncatalyzed reaction with the formula k = Ae^-Ea/RT, like how you had to solve for problem #67. In this case, you would make the kcat/kuncat equal to 1000.

- Tue Mar 13, 2018 12:29 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Hw 15.67
- Replies:
**5** - Views:
**233**

### Re: Hw 15.67

Use the equation k = Ae^(-Ea/RT). Because you're just trying to find the factor by which the catalyzed and uncatalyzed reactions differ, divide the k of the catalyzed reaction by the k of the uncatalyzed to get the term: cat refers to catalyzed; uncat refers to uncatalyzed kcat/kuncat = Ae^(-Ea,cat/...

- Tue Mar 13, 2018 12:15 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: 15.79 Kinetic Control Vs Kinetic Favorability
- Replies:
**2** - Views:
**140**

### Re: 15.79 Kinetic Control Vs Kinetic Favorability

For part (a), the second product is formed at high temperatures, so a large amount of energy is needed to overcome the energy barrier for the product to form. For (b), kinetic control predominates at lower temperatures, because the reaction pathways with lower energy barriers have larger rate consta...

- Tue Mar 06, 2018 8:51 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: 15.20
- Replies:
**1** - Views:
**66**

### Re: 15.20

a) A - 3rd order

C - 1st order

overall order = 4

b) rate = k[A]^3[C]

c) k = 0.055 L^3*mol^-3*s^-1

d) 1.80 mol*L^-1*s^-1

C - 1st order

overall order = 4

b) rate = k[A]^3[C]

c) k = 0.055 L^3*mol^-3*s^-1

d) 1.80 mol*L^-1*s^-1

- Tue Mar 06, 2018 8:36 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: 15.18
- Replies:
**1** - Views:
**54**

### Re: 15.18

These are the answers that I got:

a) A - 2nd order

B - 1st order

overall order - 3

b) rate = k[A]^2[B]

c) k = 1.2 x 10^2 L^2*mol^-2*s^-1

d) rate = 0.87 mol*L^-1*s^-1

a) A - 2nd order

B - 1st order

overall order - 3

b) rate = k[A]^2[B]

c) k = 1.2 x 10^2 L^2*mol^-2*s^-1

d) rate = 0.87 mol*L^-1*s^-1

- Tue Mar 06, 2018 8:27 pm
- Forum: First Order Reactions
- Topic: Ordered Reactions
- Replies:
**1** - Views:
**73**

### Re: Ordered Reactions

Concerning the rate or speed of the reaction,if both a first and second order reaction have substance of the same concentration, the second order reaction will have a higher rate than the first because the concentration is squared. I don't think they really are explicit differences in the way a reac...

- Sat Mar 03, 2018 11:49 am
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Overall order
- Replies:
**4** - Views:
**130**

### Re: Overall order

I actually have the same question and the answers below didn't quite help. I understand that the overall order of the reaction is the sum of the orders of the individual reactants and that these can differ. The individual orders can also help determine the rate constant. However, what is the point o...

- Wed Feb 28, 2018 7:48 am
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Rate constant
- Replies:
**2** - Views:
**88**

### Re: Rate constant

The equilibrium constant is a number that expresses the relationship between the products and reactants present at equilibrium in a chemical reaction. Typically, the concentration of the products is divided by the concentration of the reactants, and each is raised to the power of its stoichiometric ...

- Wed Feb 28, 2018 7:27 am
- Forum: General Rate Laws
- Topic: Negative in 15.5
- Replies:
**3** - Views:
**147**

### Re: Negative in 15.5

^^^The answer above is corrected. It really depends on the wording of the question. For example, the rate at which oxygen is consumed would be negative because the oxygen is consumed and thus the amount of oxygen decreases. However, if the question asked for the rate at which oxygen is consumed, the...

- Fri Feb 23, 2018 1:26 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Electromotive Force
- Replies:
**3** - Views:
**124**

### Re: Electromotive Force

According to page 572 of the textbook, it's another name for cell potential. Cell potential is the measure of the potential difference between two half cells of an electrochemical cell. When it is measured by a voltmeter without drawing current, the potential difference is the maximum potential that...

- Fri Feb 23, 2018 1:11 pm
- Forum: Balancing Redox Reactions
- Topic: H+ or H3O+ [ENDORSED]
- Replies:
**3** - Views:
**130**

### Re: H+ or H3O+ [ENDORSED]

As Lavelle pointed out it lecture today, you can use either.

Here's the post he showed in class:

viewtopic.php?f=139&t=28011

Here's the post he showed in class:

viewtopic.php?f=139&t=28011

- Sun Feb 18, 2018 5:10 pm
- Forum: Balancing Redox Reactions
- Topic: Bruincast
- Replies:
**4** - Views:
**151**

### Re: Bruincast

I'm not sure, but I think you can try this link (will include below). You just sign in using your UCLA login and can see all the recorded lectures.

https://d7.oid.ucla.edu

https://d7.oid.ucla.edu

- Tue Feb 13, 2018 12:03 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 9.81
- Replies:
**1** - Views:
**85**

### Re: 9.81

Thermodynamic stability depends on whether a reaction is spontaneous, so we can look to ΔG to figure this out. The gibbs free energy for the reaction in problem 81 is negative, so the reaction is spontaneous. Since the reaction is spontaneous in the direction of the product, the product Fe2O3 is mor...

- Tue Feb 13, 2018 11:57 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: 9.75
- Replies:
**1** - Views:
**86**

### Re: 9.75

In some cases, with simpler substances, I find it easy to draw their structures and determine the number of different orientations they can have. For #75, I looked at the given diagrams of the trans and cis molecules and figured out how many orientations each can have. For the arrangement of the cis...

- Sat Feb 10, 2018 1:39 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: When to use + sign
- Replies:
**11** - Views:
**278**

### Re: When to use + sign

I think generally using a + sign is just considered good notation. For example, if the answer's 9.12 J, it will often be written as +9.12 J.

- Fri Feb 09, 2018 2:52 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Gibbs Free Energy
- Replies:
**1** - Views:
**59**

### Re: Gibbs Free Energy

According to lecture notes, G is defined as the energy available to do work when both temperature and pressure are constant. You can see how it's derived on pages 355 and 356 of the textbook. I also found this post from 2015.

viewtopic.php?t=5141

viewtopic.php?t=5141

- Fri Feb 09, 2018 2:40 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Calculating the change in entropy with constant pressure
- Replies:
**2** - Views:
**88**

### Re: Calculating the change in entropy with constant pressure

For previous questions, C(p,m) = (5/2)R, so I assume you can just use this instead of C.

S = n(5/2)Rln(T2/T1)

S = n(5/2)Rln(T2/T1)

- Sun Feb 04, 2018 12:50 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 9.25
- Replies:
**5** - Views:
**151**

### Re: 9.25

There are 6 different orientations for SO2F2, so W = 6^N, where N is Avogadro's number. 6 is raised to Avogadro's number to account for 1 mole of the substance, rather than just 1 molecule.

- Sat Feb 03, 2018 11:02 am
- Forum: Van't Hoff Equation
- Topic: Final Van't Hoff Eqn
- Replies:
**2** - Views:
**244**

### Re: Final Van't Hoff Eqn

at T1: ln(K1) = (-ΔH°/RT1) + (ΔS°/R) at T2: ln(K2) = (-ΔH°/RT2) + (ΔS°/R) if you assume that ΔS° is constant and subtract ln(K1) from ln(K2), the (ΔS°/R) cancels out ln(K2) - ln(K1) = (-ΔH°/RT2) + (ΔS°/R) - [(-ΔH°/RT1) + (ΔS°/R)] = (-ΔH°/RT2) + (ΔS°/R) + (ΔH°/RT1) - (ΔS°/R) = (-ΔH°/RT2) + (ΔH°/RT1) ...

- Sat Feb 03, 2018 10:54 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Gibbs Free Energy Equations
- Replies:
**3** - Views:
**113**

### Re: Gibbs Free Energy Equations

Gibbs Free Energy is defined as: G = H -TS For a process at constant temperature: ΔG = ΔH - T ΔS or (ΔG-T) = (ΔH/T) - ΔS At constant temperature and pressure ΔG = -T ΔStot To find the Gibbs free energy of reaction, ΔG, using difference in molar Gibbs free energies, Gm ΔG = ∑nGm (products) - ∑nGm (re...

- Fri Jan 26, 2018 3:42 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Entropy factors [ENDORSED]
- Replies:
**2** - Views:
**95**

### Re: Entropy factors [ENDORSED]

Changes in temperature will lead to changes in entropy. The higher the temperature, the more thermal energy there is. If there's more thermal energy, there are more ways to distribute that energy, leading to a higher entropy. Thus, increasing temperature increases entropy. Similarly, increasing the ...

- Fri Jan 26, 2018 3:19 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Refresher on sig figs
- Replies:
**3** - Views:
**135**

### Re: Refresher on sig figs

Exact numbers have an infinite amount of sig figs. For addition and subtraction, however, you use the least amount of decimal places. If you need to review sig figs, you can use the doc on Lavelle's website! :) https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14B/EVERYTHING_YOU_WANTED_T...

- Fri Jan 26, 2018 3:10 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Heat Capacities of monatomic gas and linear molecule
- Replies:
**2** - Views:
**107**

### Re: Heat Capacities of monatomic gas and linear molecule

Hi, Joyce! Heat capacity is the heat required to raise the temperature of a gas by 1 degrees Celsius. dQ = nC DT The value of 3/2 R is derived from the average kinetic energy of an ideal, monatomic gas. I think the values (3/2)R and (5/2)R have to do with kinetic theory, which is something we haven'...

- Thu Jan 18, 2018 7:22 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.57
- Replies:
**3** - Views:
**145**

### Re: 8.57

I think of it as a little puzzle. Basically, you either reverse the given reactions and enthalpies or multiply them by an integer in a way that makes their sum equal to the reaction for the hydrogenation of ethyne to ethane. The given ΔHc implies combustion. C2H2(g)+ 2 H2(g) --> C2H6(g) This is the ...

- Thu Jan 18, 2018 6:36 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: 8.25
- Replies:
**2** - Views:
**120**

### Re: 8.25

So the question is asking you to find ΔU using the information provided. calibration of the calorimeter: q = -3.5- kJ, V = 0.200 L, ΔT = 7.32 degrees Celsius subsequent experiment: 100.0 mL of 0.200 M KOH, ΔT = 2.49 degrees Celsius -q = qcal and Ccal = qcal/ΔT to find the C (specific heat) from the ...

- Thu Jan 18, 2018 4:42 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: enthalpy of formation
- Replies:
**2** - Views:
**98**

### Re: enthalpy of formation

I may be wrong but I think the standard enthalpy of formation of an element in its most stable form is zero because it's defined that way. We can't measure enthalpy and can only measure changes in enthalpy, so scientists need a baseline so the enthalpy of other elements can be defined relative to th...

- Thu Jan 18, 2018 4:37 pm
- Forum: Calculating Work of Expansion
- Topic: Self-Test 8.1A
- Replies:
**1** - Views:
**97**

### Re: Self-Test 8.1A

I'm actually getting a different answer too! I was confused but I also found a post from a couple years ago. Perhaps there's a typo?

viewtopic.php?t=2486

viewtopic.php?t=2486

- Sun Jan 14, 2018 6:21 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: reversible vs irreversible
- Replies:
**2** - Views:
**169**

### Re: reversible vs irreversible

The reversible expansion of an ideal gas has a pressure that keeps changing, so you have to use an integral to find the area under the curve. The equation that is used for reversible expansion is w=-nRTln(V2/V1) The irreversible expansion is against a constant external pressure, so you don't need an...

- Fri Jan 12, 2018 1:03 pm
- Forum: Administrative Questions and Class Announcements
- Topic: HW for week 2 disc
- Replies:
**5** - Views:
**205**

### Re: HW for week 2 disc

You can choose any problems! The only requirement is that you do 7 questions per week.

- Thu Jan 11, 2018 10:34 pm
- Forum: Phase Changes & Related Calculations
- Topic: constant pressure
- Replies:
**3** - Views:
**136**

### Re: constant pressure

Here is a somewhat long explanation of how I interpreted it. Let's consider gas in a piston. When the gas is heated, the kinetic energy of the gas molecules increases, and the individual molecules move faster, colliding more with the piston. The collisions with the piston transfer the energy to the ...

- Thu Dec 07, 2017 10:25 am
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Equilibria shift with temp. increase 11.77
- Replies:
**2** - Views:
**143**

### Re: Equilibria shift with temp. increase 11.77

I like to think of heat as either a reactant or a product. So in an exothermic reaction, where heat is given off, heat can be thought of as a product. Thus, increasing heat will increase the amount of product, and the reaction will favor the reactants. Similarly, in an endothermic reaction, heat is ...

- Thu Dec 07, 2017 10:20 am
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: How to tell if bidentate, tri- etc
- Replies:
**5** - Views:
**330**

### Re: How to tell if bidentate, tri- etc

Polydentate refers to a ligand that can bind the same transition metal more than once. Chelate refers to a ring-like structure involving a transition metal and a ligand. These structures result from a polydentate ligand binding the same transition metal.

- Wed Nov 29, 2017 7:49 pm
- Forum: Hybridization
- Topic: Delocalized π bonds
- Replies:
**2** - Views:
**176**

### Re: Delocalized π bonds

I like to think of delocalized π bonds in terms of the electrons in the bonds. You can also think about the electrons being delocalized, meaning that the electrons within the bonds can exist in different conformations, similar to having resonant structures.

- Wed Nov 29, 2017 7:32 pm
- Forum: Hybridization
- Topic: 4.31
- Replies:
**2** - Views:
**148**

### Re: 4.31

You can use regions of electron density to easily determine shape and hybridization and vice-versa. For example, 3 regions of electron density corresponds to both a trigonal planar shape and sp^2 hybridization (s p p). 4 regions of electron density corresponds to a tetrahedral shape sp^3 hybridizati...

- Tue Nov 21, 2017 3:49 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Bent vs. Angular
- Replies:
**5** - Views:
**260**

### Re: Bent vs. Angular

There's no difference. Both bent and angular refer to the shape of AX2E or AX2E2 molecules

- Tue Nov 21, 2017 3:44 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Dipole Moments
- Replies:
**1** - Views:
**86**

### Re: Dipole Moments

I think we use the original convention...the examples shown in class and in my discussion sessions have showed arrows pointing toward the more electronegative.

- Sun Nov 19, 2017 10:27 pm
- Forum: Dipole Moments
- Topic: Question 4.29
- Replies:
**2** - Views:
**144**

### Re: Question 4.29

The dipole moment for 1 would be largest because the bond vectors for C-Cl are pointing in almost the same direction but in 2 the vectors point in different ways, which cancels out the dipole.

- Sun Nov 19, 2017 10:23 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Naming; VSEPR
- Replies:
**3** - Views:
**171**

### Re: Naming; VSEPR

A is used to denote the central atom

X denotes the bonded atom

E denotes the lone pair

X denotes the bonded atom

E denotes the lone pair

- Wed Nov 08, 2017 9:41 am
- Forum: Electronegativity
- Topic: 3.79
- Replies:
**3** - Views:
**231**

### Re: 3.79

You can use the difference in electronegativity to figure this out. If the electronegativity difference > 2, then it's an ionic bond. If the electronegativity difference < 1.5, then it's a covalent bond. Figure 3.12 on page 92 of the textbook gives the electronegativities of the main group elements....

- Wed Nov 08, 2017 9:20 am
- Forum: Ionic & Covalent Bonds
- Topic: 3.25
- Replies:
**3** - Views:
**163**

### Re: 3.25

I like to think of it as how many electrons the elements in a group need to gain or lose to have the electron configuration of a noble gas. For example, the elements in Group 17 (F, Cl, Br, etc) need to gain 1 electron to have the configuration of a noble gas, so they have a charge of -1. The elemen...

- Wed Nov 01, 2017 8:42 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: Post-Module #23 and Textbook number 1.45
- Replies:
**1** - Views:
**130**

### Re: Post-Module #23 and Textbook number 1.45

The uncertainty in velocity for #45 should be 10 m/s. The final answer should be 6.7 x 10^-37 m. This problem is actually noted in the "Solutions Manual Errors" part of his website.

- Wed Nov 01, 2017 8:36 pm
- Forum: Lewis Structures
- Topic: Electron Configurations
- Replies:
**2** - Views:
**125**

### Re: Electron Configurations

The answer to part (a) is Co3+ because Co has the electron configuration: [Ar] 3d7 4s2 However, Co3+ loses 3 electrons (2 from 4s and 1 from the 3d orbitals). The new configuration is [Ar] 3d6 Similarly, for part (b), if two electrons from the 4s orbital and one electron from the 3d orbital are lost...

- Thu Oct 26, 2017 10:44 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Ch.3 #5
- Replies:
**1** - Views:
**131**

### Re: Ch.3 #5

The ground-state electron configuration for Cu+ would be [Ar] 3d10.

- Wed Oct 25, 2017 9:59 pm
- Forum: Quantum Numbers and The H-Atom
- Topic: Values of Quantum Numbers
- Replies:
**3** - Views:
**212**

### Re: Values of Quantum Numbers

This is #19 from Chapter 2, right? Refer to Table 2.2 on page 36 of the textbook a) 7 values there are n different values of l for a given value of n; for example, when n=3, l can have any of the three values 0, 1, and 2 because l has the value of n-1, so it can be 0, 1, 2, 3, 4, 5, or 6 b) 5 values...

- Tue Oct 17, 2017 1:33 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Module- definition of spectroscopic experiment?
- Replies:
**2** - Views:
**291**

### Re: Module- definition of spectroscopic experiment?

I believe the correct answer is A

- Tue Oct 17, 2017 1:32 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Atomic Spectra Post-Module #28
- Replies:
**1** - Views:
**128**

### Re: Atomic Spectra Post-Module #28

To find the wavelength with the given information, divide 1m by 1,650,763.73 wavelengths. The length of a single wave is 6.058 x 10^-7 m or 605.8 nm This wavelength corresponds to visible light, which has the range of about 400-700 nm. To find the energy of one photon, use the formula E = hc/lambda ...

- Thu Oct 12, 2017 11:22 pm
- Forum: Einstein Equation
- Topic: Chapter 1 #23
- Replies:
**6** - Views:
**419**

### Re: Chapter 1 #23

Yes, you must convert to joules! It has to do with the cancellation of units. I just looked it up on google 1eV = 1.602 x 10^-19 J Convert the energy, and you get 2.2510 x 10^-14 J If you use the formula lambda =hc/E, the joules cancel out. I will show you the solution below... lambda = [(6.626 x 10...

- Thu Oct 12, 2017 11:15 pm
- Forum: Photoelectric Effect
- Topic: Chapter 1, #27
- Replies:
**1** - Views:
**152**

### Re: Chapter 1, #27

I also started by finding the total energy; however, I used the formula E=hc/lambda. The given wavelength is in nm so convert to m. E = [(6.626 x 10^-34 Js)(3.0 x 10^8 m/s)]/(420 x 10^-9 m) = 4.7 x 10^-19 J Doing this gives us how much energy in Joules each photon has. Then, the question tells us th...

- Thu Oct 12, 2017 11:05 pm
- Forum: Einstein Equation
- Topic: Lyman and Blamer series [ENDORSED]
- Replies:
**7** - Views:
**551**

### Re: Lyman and Blamer series [ENDORSED]

Basically, the initial energy levels for the series are different. For the Balmer series, n1=2, but for the Lyman series, n1=1.

- Thu Oct 05, 2017 11:06 pm
- Forum: Limiting Reactant Calculations
- Topic: M11
- Replies:
**1** - Views:
**168**

### Re: M11

You have to find out the number of moles of O2 used in the first reaction. Start with 5.77 grams of P4, convert to this to moles, and use the equation to find moles of O2. Then convert the moles of O2 found to grams. From this, you know that 4.47 grams of O2 were used in the first reaction. Then, co...

- Thu Oct 05, 2017 10:40 pm
- Forum: Accuracy, Precision, Mole, Other Definitions
- Topic: Formula Units
- Replies:
**2** - Views:
**186**

### Re: Formula Units

A formula unit is a group of ions. Formula units are different for each ionic compound. For example, the formula unit of (NH4)2SO4 consists of two NH4^+ ions and one SO4^2- ion. But, NaCl consists of one Na^+ ion and one Cl^- ion.

1 mole = 6.022 x 10^23 formula units.

I hope this cleared it up a bit!

1 mole = 6.022 x 10^23 formula units.

I hope this cleared it up a bit!

- Thu Oct 05, 2017 10:31 pm
- Forum: Limiting Reactant Calculations
- Topic: M3
- Replies:
**3** - Views:
**389**

### Re: M3

So, the question gives us the actual yield, which is 17.5 g of carbon dioxide. However, I don't think you need to know if CaCO3 is the limiting reactant... Using the given mass of CaCO3, you can convert this to moles then use the stoichiometric coefficients in the equation to find moles of CO2 and t...