Search found 66 matches
- Sat Mar 17, 2018 12:42 am
- Forum: Zero Order Reactions
- Topic: graph of 0 order
- Replies: 10
- Views: 1954
Re: graph of 0 order
The graph would be a horizontal line that abruptly drops straight down to zero (once all of the reactants are used up)
- Sat Mar 17, 2018 12:41 am
- Forum: Limiting Reactant Calculations
- Topic: Reactant vs. Reagent [ENDORSED]
- Replies: 23
- Views: 34420
Re: Reactant vs. Reagent [ENDORSED]
Used interchangeably
- Sat Mar 17, 2018 12:39 am
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Homework Problem 14.27
- Replies: 2
- Views: 578
Re: Homework Problem 14.27
E is not a state function, so you can't simply add them together. G is a state function, so that is why we must convert to G.
- Tue Mar 13, 2018 4:26 pm
- Forum: Balancing Redox Reactions
- Topic: Test 2 #1
- Replies: 6
- Views: 839
Re: Test 2 #1
O2 has a charge of 0. For CO2, the O has a charge of -2, and since there are 2 "O"s in CO2 the charge for O is -4. In order to cancel this -4 out, the charge on C must be +4. O2 is reduced since its charge goes from 0 to -4, and thus gained electrons.
- Tue Mar 13, 2018 4:23 pm
- Forum: Second Order Reactions
- Topic: Half life
- Replies: 6
- Views: 938
Re: Half life
If you plug it in to the equation, you should get the correct answer. 1/(1/16) is 16.
- Tue Mar 13, 2018 4:20 pm
- Forum: Zero Order Reactions
- Topic: Chart problems [ENDORSED]
- Replies: 3
- Views: 625
- Thu Mar 08, 2018 4:23 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Units for rate
- Replies: 4
- Views: 659
Re: Units for rate
It is okay to leave your answer in mmol as long as the question doesn't ask for the answer in mol
- Thu Mar 08, 2018 4:22 pm
- Forum: First Order Reactions
- Topic: One of the equations for 1st order rxn
- Replies: 3
- Views: 544
Re: One of the equations for 1st order rxn
d[A]/[A] stands for the rate of consumption of [A] during a first order reaction.
- Thu Mar 08, 2018 4:20 pm
- Forum: Second Order Reactions
- Topic: 15.27 vs 15.35
- Replies: 3
- Views: 771
Re: 15.27 vs 15.35
This is because in second order reactions, half life depends on initial concentration, whereas half life in first order reactions does not depend on concentration.
- Thu Mar 08, 2018 4:19 pm
- Forum: First Order Reactions
- Topic: Equation variations
- Replies: 9
- Views: 1188
Re: Equation variations
^ They just rearranged the equation. Both versions should yield the same answer
- Thu Mar 08, 2018 4:19 pm
- Forum: First Order Reactions
- Topic: Equation variations
- Replies: 9
- Views: 1188
Re: Equation variations
Yes, all they did in the book was switch the ln[A]f to the right side, and moved -kt to the left side. ln[A]o-ln[A]f is the same as ln([A]o/[A]f)
- Thu Mar 08, 2018 4:15 pm
- Forum: Zero Order Reactions
- Topic: zero order rate?
- Replies: 14
- Views: 1556
Re: zero order rate?
True, in zero order reactions the rate does not depend on concentration. d[A]=-kdt
- Thu Mar 08, 2018 4:14 pm
- Forum: Second Order Reactions
- Topic: 15.35
- Replies: 3
- Views: 655
Re: 15.35
For second order half life, the half life depends on the initial concentration, t(1/2)=1/(k[A]o) whereas in first order reactions, half life only depends on k
- Thu Mar 08, 2018 4:12 pm
- Forum: General Rate Laws
- Topic: 15.17 type problems
- Replies: 4
- Views: 692
Re: 15.17 type problems
For work, I usually show something like 2^a=2 (the concentration doubled, rate doubled), therefore a=1, and the order of [A] is first order. But you need to show work, at least on the quizzes and tests,
- Thu Mar 08, 2018 4:10 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: 15.17
- Replies: 6
- Views: 1048
Re: 15.17
Yes, once C is determined to be zero order, you can ignore the concentrations for C entirely, since it does not have an effect on the reaction rate.
- Wed Feb 28, 2018 10:18 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Kinetics Test Week 9
- Replies: 2
- Views: 473
Re: Kinetics Test Week 9
The test will cover 15.1-15.6, which is up to #39 in the homework problems.
- Wed Feb 28, 2018 10:17 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Test 3
- Replies: 6
- Views: 864
Re: Test 3
Up to #39, I checked with a TA!
- Wed Feb 28, 2018 10:15 pm
- Forum: General Rate Laws
- Topic: Overall order [ENDORSED]
- Replies: 6
- Views: 974
Re: Overall order [ENDORSED]
After figuring out the order of each reactant, you just add the orders together to get the overall order
- Wed Feb 28, 2018 10:13 pm
- Forum: Zero Order Reactions
- Topic: Units [ENDORSED]
- Replies: 7
- Views: 1880
Re: Units [ENDORSED]
Yes! M/s
- Mon Feb 26, 2018 4:29 pm
- Forum: General Rate Laws
- Topic: Negative Sign [ENDORSED]
- Replies: 7
- Views: 917
Re: Negative Sign [ENDORSED]
In order to make the rate positive, the negative sign reverses the negative rate of the product being used up in a reaction.
- Mon Feb 26, 2018 4:26 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Midterm Question 7 Sig Figs
- Replies: 2
- Views: 2754
Re: Midterm Question 7 Sig Figs
Usually you wait until the end to round your answers, but you should definitely ask your TA about it.
- Mon Feb 26, 2018 4:25 pm
- Forum: Zero Order Reactions
- Topic: Zero Order Meaning
- Replies: 5
- Views: 1237
Re: Zero Order Meaning
Rate is equal to k in a zero order reaction, and therefore plotting the points creates a straight line
- Mon Feb 26, 2018 4:24 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Q and K
- Replies: 7
- Views: 849
Re: Q and K
Q and K are constant throughout chemistry, having mainly to do with equilibrium.
- Mon Feb 26, 2018 4:21 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 3000768
Re: Post All Chemistry Jokes Here
what kind of dogs do chemists have?
Labratory retrievers
Labratory retrievers
- Sun Feb 25, 2018 4:19 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Determining N
- Replies: 4
- Views: 597
Re: Determining N
No it does not matter, "n" will be the same no matter what.
- Sun Feb 25, 2018 4:17 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 3000768
Re: Post All Chemistry Jokes Here
I don't trust atoms....
I heard they make up everything
I heard they make up everything
- Sun Feb 25, 2018 4:16 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: swapping signs of E values
- Replies: 8
- Views: 3593
Re: swapping signs of E values
I use Eo(cell) = Eo(cathode) - Eo(anode) so that way I do not need to flip any signs.
- Wed Feb 14, 2018 8:39 pm
- Forum: Van't Hoff Equation
- Topic: Practice Test
- Replies: 5
- Views: 1068
Re: Practice Test
Nope!
- Wed Feb 14, 2018 8:39 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: adibatic [ENDORSED]
- Replies: 2
- Views: 536
Re: adibatic [ENDORSED]
Adiabatic only implies that q=0, not delta T
- Wed Feb 14, 2018 8:37 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: PV = nRT
- Replies: 6
- Views: 940
Re: PV = nRT
For ideal gases when you need to solve for a constant.
- Sat Feb 10, 2018 8:47 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Practice Midterm Winter 2018
- Replies: 10
- Views: 3112
Re: Practice Midterm Winter 2018
Where do I find the link?
- Fri Feb 09, 2018 4:25 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Question 9.55
- Replies: 2
- Views: 399
Re: Question 9.55
You do. It isn't shown in the solution manual because the reactants in these problems have an Hf of 0 (for example, O2, H2, C(S) all have an Hf=0). Therefore you only need to look at the Hf of the product side (in this case).
- Fri Feb 09, 2018 4:21 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 9. 43
- Replies: 5
- Views: 583
Re: 9. 43
You won't have to calculate it, it should be given in the problem. They make you look it up in the book for the homework problems, but on the test it will be given.
- Tue Jan 30, 2018 9:00 am
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Test question [ENDORSED]
- Replies: 16
- Views: 1698
Re: Test question [ENDORSED]
I stated that energy cannot be created nor destroyed, only transferred. I'm pretty sure that is all they would be looking for, because the First Law of Conservation is primarily concerned with the conservation of energy.
- Tue Jan 30, 2018 8:58 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: unit of entropy
- Replies: 11
- Views: 1173
Re: unit of entropy
Joules per Kelvin (J/K)
- Tue Jan 30, 2018 8:58 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: A spontaneous reaction
- Replies: 7
- Views: 917
Re: A spontaneous reaction
Spontaneous reactions occur without any type of outside intervention/force. Spontaneous reactions favor the formation of products at the current conditions under which the reaction is occurring.
- Tue Jan 23, 2018 6:42 pm
- Forum: Phase Changes & Related Calculations
- Topic: HW 8.31
- Replies: 3
- Views: 462
Re: HW 8.31
For this question, you would use q=mC(delta)T. For part a, where the pressure is held constant (Cp), C would be equal to (5/2)R, where R=8.314. In part b, where the volume is held constant (Cv), C would be equal to (3/2)R, where R is also 8.314. So basically, you would just replace the "C"...
- Tue Jan 23, 2018 6:38 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Hess's Law
- Replies: 5
- Views: 867
Re: Hess's Law
The molecules must be on opposite sides and I=equal in quantity to be able to fully cancel them out. For example, if there are 2 modes of CO2 on the reactant side of one equation ands you are adding this equation to one in which there are 2 moles of CO2 on the product side, the CO2's will cancel out.
- Tue Jan 23, 2018 6:36 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Hess' Law fractions
- Replies: 3
- Views: 1171
Re: Hess' Law fractions
Since enthalpy is a state function, it does not follow a specific pathway, and thus it it okay to use any coefficients as long as you consistently change the enthalpy along with it.
- Wed Jan 17, 2018 5:45 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Hw 8.67 part b
- Replies: 2
- Views: 321
Re: Hw 8.67 part b
717 kJ/mol would have to be given. It can be found in the appendix at the end of the book.
- Wed Jan 17, 2018 5:40 pm
- Forum: Phase Changes & Related Calculations
- Topic: 8.5
- Replies: 5
- Views: 472
Re: 8.5
Yes, because the cylinder is being supplied with energy, this energy (524) will be positive. In addition, because work is being done on the system by the surroundings, 340 will be positive as well.
- Wed Jan 17, 2018 5:37 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Q 8.13
- Replies: 9
- Views: 978
Re: Q 8.13
If the system is doing work on its surroundings, as in this problem, the work is negative.
- Sat Jan 13, 2018 12:21 pm
- Forum: Phase Changes & Related Calculations
- Topic: Heat and work not state functions
- Replies: 7
- Views: 885
Re: Heat and work not state functions
I found this online and found it to be very helpful. "Heat and work, unlike temperature, pressure, and volume, are not intrinsic properties of a system. They have meaning only as they describe the transfer of energy into or out of a system. "Heat & work are the forms of energy in trans...
- Sat Jan 13, 2018 12:17 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Specific heat capacity vs molar heat capacity
- Replies: 5
- Views: 588
- Sat Jan 13, 2018 12:15 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Heat Not a State Property
- Replies: 6
- Views: 512
Re: Heat Not a State Property
Hi, I found this online and found it very helpful. "Heat and work, unlike temperature, pressure, and volume, are not intrinsic properties of a system. They have meaning only as they describe the transfer of energy into or out of a system." "They appear only when there occurs any chang...
- Fri Dec 08, 2017 8:12 pm
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: 12.29
- Replies: 1
- Views: 416
Re: 12.29
It depends on what is produced. If H3O+ is produced in the reaction, then you can solve for pH. If OH- is produced, then you can solve for pOH.
- Fri Dec 08, 2017 8:10 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: 12.55/57
- Replies: 2
- Views: 456
Re: 12.55/57
You use an ICE table when a weak acid or base is being reacted with water. A strong acid or base completely dissociates in water, so you can go straight to the equilibrium constant expression.
- Tue Nov 28, 2017 5:11 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Coordination compounds and complexes
- Replies: 2
- Views: 261
Re: Coordination compounds and complexes
A coordination compound is the result of a Lewis acid-base reaction in which ligands (anions) bond to the central metal atom, or ion. Ligands are Lewis bases, meaning they have an extra pair to pairs of electrons to donate to the Lewis acid (the central metal atom).
- Tue Nov 28, 2017 5:08 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: SF6
- Replies: 1
- Views: 281
Re: SF6
All of the bonds are 90 degrees. It says 180 for the two atoms directly across from one another.
- Tue Nov 21, 2017 5:23 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 11.79
- Replies: 2
- Views: 301
Re: 11.79
For this problem, we would need to use a graphing calculator in order to solve for "x". We shouldn't get a problem like this on a test unless we are allowed to use a graphing calculator.
- Tue Nov 21, 2017 5:20 pm
- Forum: Ideal Gases
- Topic: Q and K [ENDORSED]
- Replies: 35
- Views: 3014
Re: Q and K [ENDORSED]
The two formulas are the same. You compare Q to K in order to determine which way the reaction will shift (sit). If Q<K, it will shift/sit right (toward products), if Q>K, it will shift/sit left (towards reactants).
- Sat Nov 18, 2017 6:52 pm
- Forum: Bond Lengths & Energies
- Topic: Bond length
- Replies: 2
- Views: 318
Re: Bond length
I believe the length would be closer to a double bond, but in between the lengths of a single and double bond.
- Sat Nov 18, 2017 6:50 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Bent or linear?
- Replies: 11
- Views: 1984
Re: Bent or linear?
The shape depends on the number of lone pairs as well as their placement on the molecule.
- Sun Nov 12, 2017 4:10 pm
- Forum: Polarisability of Anions, The Polarizing Power of Cations
- Topic: Polar vs nonpolar vs ionic
- Replies: 11
- Views: 28282
Re: Polar vs nonpolar vs ionic
Another thing to note for polarity is in regard to the dipole moments. If the dipole moments are equal and cancel each other out, the molecule is non polar; however, if the dipole moments do not cancel each other out, the molecule is polar.
- Sun Nov 12, 2017 4:05 pm
- Forum: Dipole Moments
- Topic: Dipole moments
- Replies: 6
- Views: 1097
Re: Dipole moments
In addition to what has already been said, if the dipole moments around a molecule are equal (cancel each other out), the molecule is non polar. If the dipole moments are not equal (do not cancel each other out), the molecule is polar.
- Fri Nov 03, 2017 6:06 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: 3D shape Lewis Diagrams
- Replies: 4
- Views: 841
Re: 3D shape Lewis Diagrams
@Swetha No, I don't believe so.
- Fri Nov 03, 2017 6:05 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Midterm
- Replies: 8
- Views: 893
Re: Midterm
On the exam and homework schedule, it says that the midterm covers all material up to November 3, so I think it might be. I can ask Lavelle.
- Wed Oct 25, 2017 3:15 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: 2.75
- Replies: 2
- Views: 336
Re: 2.75
Since the s-block has lower ionization energies than the p-block, it takes less energy to remove an electron from the s-block than the p-block. Because it can lose electrons easier, the s-block is thus more reactive than the p-block.
- Wed Oct 25, 2017 3:12 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: #55 d
- Replies: 3
- Views: 407
Re: #55 d
A half filled D-orbital (d^5) or a filled D-orbital (d^10) is more stable than d^4 or d^9, for example, so the electron from the S-orbital will "jump" to the d-orbital in order to make it more stable. In this case, the electron will jump to the d-orbital in order to completely fill it (d^1...
- Thu Oct 19, 2017 2:23 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: Upcoming Test
- Replies: 5
- Views: 536
Re: Upcoming Test
Hey, I've already taken the test and you do not need to know about Heisenberg's uncertainty principle for this test. You are only being tested on Chapter 1 sections 1-5. As for the homework, my TA had us turn in 4 problems that we used to study for the test (so I turned in 4 problems from sections 1...
- Tue Oct 17, 2017 1:21 pm
- Forum: Properties of Light
- Topic: Problem 1.23
- Replies: 3
- Views: 499
Re: Problem 1.23
The conversion between KeV and Joules will be given either in the problem or on the conversion/formula sheet. We are not expected to memorize the conversion.
1 eV = 1.6022 x 10^-19 J
1 eV = 1.6022 x 10^-19 J
- Wed Oct 11, 2017 10:20 am
- Forum: Photoelectric Effect
- Topic: Chapter 1, Question 1.23
- Replies: 3
- Views: 423
Re: Chapter 1, Question 1.23
First, you must convert 140.511 eV to Joules. (1eV = 1.6022 x 10^-19 J).
From there, you use the equation E=hc/lambda and solve for lambda (wavelength).
You should get 8.8237 pm for your answer.
From there, you use the equation E=hc/lambda and solve for lambda (wavelength).
You should get 8.8237 pm for your answer.
- Wed Oct 11, 2017 10:16 am
- Forum: Properties of Light
- Topic: Chapter 1: Exercise 15
- Replies: 4
- Views: 870
Re: Chapter 1: Exercise 15
In order to solve the problem, you do use the wavelength. In order to solve for n2, you must first find the frequency (v) so that you can plug the frequency into Rydberg's equation. To find the frequency, you use the equation v=c/lambda. Then you use plug this frequency into Rydberg's equation, v=R(...
- Thu Oct 05, 2017 12:04 am
- Forum: Accuracy, Precision, Mole, Other Definitions
- Topic: E.29 part b HELP
- Replies: 3
- Views: 453
Re: E.29 part b HELP
Since you are solving for moles of Cl- ions, you have to convert the 0.0417 mol CuCl2 x 4H2O to moles of Cl-. Since one mole of CuCl2 x 4H2O contains 2 Cl- ions, you take the 0.0417 mol CuCl2 x 4H2O and multiply it by (2 mol Cl- / 1 mol CuCl2 x 4H2O). This is what my work looks like: 0.0417 mol CuCl...
- Wed Oct 04, 2017 11:54 pm
- Forum: Empirical & Molecular Formulas
- Topic: Empirical Coefficients [ENDORSED]
- Replies: 12
- Views: 5702
Re: Empirical Coefficients [ENDORSED]
Typically, you will get numbers that are very close to a whole number such as 1.97 or 3.08 which you would round to 2 and 3. In many of the problems I have done so far that I've had to multiply in order to get a whole number, I've gotten numbers like 2.67, 1.33 and 3.75 which can easily be multiplie...
- Wed Oct 04, 2017 11:47 pm
- Forum: Significant Figures
- Topic: Number of sig figs in answer
- Replies: 8
- Views: 1157
Re: Number of sig figs in answer
As long as the significant figures in your answer match the significant figures given in the problem (or whichever number you are using for calculations), you will be fine. You will get a much more accurate answer if you don't round numbers during your calculations. Just make sure that the significa...