Search found 62 matches
- Fri Mar 16, 2018 10:30 am
- Forum: General Science Questions
- Topic: Class [ENDORSED]
- Replies: 3
- Views: 711
- Wed Mar 14, 2018 12:18 am
- Forum: Second Order Reactions
- Topic: Test 3 #8
- Replies: 2
- Views: 543
Test 3 #8
8.) In a 2nd order reaction, a reactant has a known half-life of 3.00 minutes and an initial concentration of 5.00 M. How much time (in seconds) will it take for the concentration of this reactant to drop to 2.00 M?
Could someone explain how they did this question?
Could someone explain how they did this question?
- Wed Mar 14, 2018 12:15 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Midterm Question 6B
- Replies: 1
- Views: 426
Re: Midterm Question 6B
For this question I used the equation deltaS=nCvln(T2/T1) You need to find the moles first, however using PV=nRT. Everything is given besides moles, so you can solve for n: n=(1.00 atm)(53.48L)/(.08206)(303K) to get n=2.15 moles. You can then use n to solve for change in entropy, using Cv as 3/2R be...
- Wed Mar 14, 2018 12:11 am
- Forum: Phase Changes & Related Calculations
- Topic: Midterm, Question 4B
- Replies: 2
- Views: 452
Re: Midterm, Question 4B
For this question I used deltaU=q+w Since the internal energy of a system decreased by 1763 kJ, deltaU=-1763. It also gives you the value for w, saying that the system did 1357 kJ of work. When a system does work its negative. so w=-1357. From there you can solve for q, and you should get q= -406 kJ...
- Mon Mar 05, 2018 11:17 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Reaction Mechanisms on test 3?
- Replies: 2
- Views: 337
Re: Reaction Mechanisms on test 3?
no the test only covers 15.1-15.6
- Mon Mar 05, 2018 11:14 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Kinetics Test
- Replies: 2
- Views: 543
Re: Kinetics Test
My TA told us that test 3 would cover #1-39 from chapter 15
- Mon Mar 05, 2018 3:06 pm
- Forum: General Rate Laws
- Topic: Test #3 Problems
- Replies: 3
- Views: 541
Re: Test #3 Problems
my TA said it was #1-39
- Mon Mar 05, 2018 2:29 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Test 3 - Chapter 15 homework
- Replies: 3
- Views: 520
Re: Test 3 - Chapter 15 homework
my TA told us questions #1-39 will be covered on test 3!
- Mon Mar 05, 2018 2:27 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Test 2
- Replies: 4
- Views: 649
Re: Test 2
You would have to find delta G for both half reactions and then add them to find delta G of the entire reaction. Once you have that you can solve for E (standard potential of cell). This problem is very similar to 14.27 if you have the solutions manual
- Mon Mar 05, 2018 2:22 pm
- Forum: General Rate Laws
- Topic: Average rate of consumption
- Replies: 2
- Views: 2078
Average rate of consumption
When finding average rate of consumption of R, I know the equation is (- delta[R]/delta t) because the concentration of R would be decreasing during the reaction. But when stating the rate, is it always positive? Textbook question 15.3 part a has the rate of consumption of NO2 as a positive rate so ...
- Thu Mar 01, 2018 12:12 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: 15.15 [ENDORSED]
- Replies: 5
- Views: 748
Re: 15.15 [ENDORSED]
Both of the reactants are first order because their concentrations are directly proportional to the rate increase factor.
- Thu Mar 01, 2018 8:56 am
- Forum: General Rate Laws
- Topic: half life vs rate
- Replies: 2
- Views: 413
Re: half life vs rate
The normal rate of the reaction is the concentration of a reactant or product to be consumed or formed over a given time unit.
Half-life of a reaction is the amount of time needed for the reactant concentration to decrease by half its original concentration.
hope this helps!
Half-life of a reaction is the amount of time needed for the reactant concentration to decrease by half its original concentration.
hope this helps!
- Sat Feb 24, 2018 6:30 pm
- Forum: General Rate Laws
- Topic: Units for 15.3 Part B
- Replies: 4
- Views: 696
Re: Units for 15.3 Part B
I think they put that just to distinguish that they were referring to moles of oxygen. I don't think we would have to put that in every problem
- Thu Feb 22, 2018 10:12 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Conducting Metal
- Replies: 1
- Views: 278
Re: Conducting Metal
Generally if you have a metal solid in your reaction you don't need to add the Pt(s). If you only have elements in the (aq) or (l) state you must add the metal conductor so electron transfer can occur.
- Thu Feb 22, 2018 10:03 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.55
- Replies: 2
- Views: 436
Re: 14.55
You don't have to switch the sign of Eº if when calculating Eºcell you always use Eºcathode-Eºanode.
If you prefer to use the method where you add the Eºcathode and Eºanode you must flip the Eºanode.
If you prefer to use the method where you add the Eºcathode and Eºanode you must flip the Eºanode.
- Tue Feb 13, 2018 11:19 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 11.15
- Replies: 2
- Views: 369
Re: 11.15
the question states that the reaction is at 1200K
- Mon Feb 12, 2018 11:21 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Homework
- Replies: 1
- Views: 294
Re: Homework
yeah every week
- Mon Feb 12, 2018 11:20 pm
- Forum: Calculating Work of Expansion
- Topic: Reversible vs irreversible
- Replies: 4
- Views: 437
Re: Reversible vs irreversible
If a reaction goes in both directions, it is at equilibrium, and therefore it is reversible.
- Mon Feb 12, 2018 11:16 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Test 1 Question 1c
- Replies: 3
- Views: 383
Re: Test 1 Question 1c
Bond enthalpies are the least accurate way to calculate enthalpy of formation because they are averages
- Mon Feb 12, 2018 11:14 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Midterm
- Replies: 2
- Views: 460
Re: Midterm
He said we won't have to specifically derive any formulas but need to know how the formulas relate to each other and be able to substitute parts of equations when necessary
- Mon Feb 12, 2018 4:33 pm
- Forum: Administrative Questions and Class Announcements
- Topic: No class on Wednesday?
- Replies: 2
- Views: 528
Re: No class on Wednesday?
yes, Dr. Lavelle announced in lecture today that there will be no class on Wednesday.
- Mon Feb 12, 2018 4:33 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Practice Midterm Karen Leung #8 [ENDORSED]
- Replies: 4
- Views: 696
Re: Practice Midterm Karen Leung #8 [ENDORSED]
For part b you calculate the delta S for changing temperature and then add it to your answer from part a.
so you do deltaS=nRln(t2/t1)
n= moles of both the argon gas and the neon gas
so you do deltaS=nRln(t2/t1)
n= moles of both the argon gas and the neon gas
- Sat Feb 10, 2018 6:53 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: Problem 9.75
- Replies: 1
- Views: 395
Re: Problem 9.75
I asked about this at a step-up session and the UA said we would not be expected to know that!
- Sat Feb 10, 2018 6:50 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 9.85
- Replies: 1
- Views: 289
Re: 9.85
For part C, it asks if the entropy of the system is primarily a result of changes in positional disorder or thermal disorder. The system here is the solution, so if you imagine potassium nitrate dissolving in water it makes logical sense that the entropy, or disorder of the system is mostly the pota...
- Mon Feb 05, 2018 9:51 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Chapter 11 #17
- Replies: 2
- Views: 291
Re: Chapter 11 #17
I got this too. I think it's a mistake in the answer key
- Mon Feb 05, 2018 9:49 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 9.63
- Replies: 4
- Views: 466
Re: 9.63
Thermodynamic stability occurs when a system is in its lowest energy state. Therefore, compounds with positive free energy are said to be unstable while compounds with negative free energy are said to be stable.
- Mon Feb 05, 2018 9:46 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 11.17
- Replies: 10
- Views: 1065
Re: 11.17
I also got -2.74 kJ.mol-1. I think the there's a mistake in the answer key
- Mon Feb 05, 2018 1:29 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 9.21
- Replies: 3
- Views: 447
9.21
In #21, where do they get the value of Kb? Also can someone explain this question I'm confused what we are calculating
- Thu Feb 01, 2018 3:11 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Question 8.13
- Replies: 3
- Views: 399
Re: Question 8.13
In addition, when a system is irreversible, we can see it as expanding into a vacuum. This means that the ∆Ssurroundings will be equal to zero because there is no change of entropy in a pressure-less reservoir. However, this is very different from a reversible system, for which the ∆Stotal of the un...
- Fri Jan 26, 2018 10:19 am
- Forum: Phase Changes & Related Calculations
- Topic: Homework Exercise 8.43 and Phase Changes
- Replies: 1
- Views: 201
Re: Homework Exercise 8.43 and Phase Changes
When looking at a heating curve, you have heat supplied on the x axis and temperature on the y axis. The plateau's represent the phase change. The first plateau is solid changing to liquid, and the second plateau is liquid changing to gas. Notice there is no temperature change during the phase chang...
- Fri Jan 26, 2018 10:09 am
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Iron Rusting: Exo or Endo? [ENDORSED]
- Replies: 4
- Views: 12822
Re: Iron Rusting: Exo or Endo? [ENDORSED]
I think iron rusting would be exothermic since it gives off heat
- Thu Jan 25, 2018 3:05 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: The sign of q
- Replies: 5
- Views: 1177
Re: The sign of q
q is postitive is heat is added to the system and q is negative is heat is removed from the system.
- Mon Jan 22, 2018 9:13 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Question 85
- Replies: 1
- Views: 260
Re: Question 85
For part B, you must use the gas law, PV=nRT. In this problem we are given everything except for moles, so you can rearrange the equation to look like n=PV/RT. Then you can plug the given numbers in. n=(1.00 atm)(5.45 L)/ (0.08206)(273K) n=0.243 mol Now that you have moles of Nitrogen, you can multi...
- Sat Jan 20, 2018 5:21 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.51 [ENDORSED]
- Replies: 2
- Views: 350
Re: 8.51 [ENDORSED]
the standard enthalpies of formation should be in the back of the textbook in appendix 2A
- Sat Jan 20, 2018 5:10 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: 8.49
- Replies: 2
- Views: 307
8.49
In #49, how do you know that PΔV is negative?
- Wed Jan 17, 2018 10:12 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.61
- Replies: 3
- Views: 462
Re: 8.61
hi, I was confused about this too but a TA explained it to me. Since there is no ammonia (NH3) in the desired reaction, the ammonia from the reactions with known enthalpies must be cancelled out. To do this, the ammonia must be on different sides of the two reactions it appears in. Since H2 appears ...
- Tue Jan 16, 2018 11:41 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Hess's Law
- Replies: 3
- Views: 336
Re: Hess's Law
Hess's law states that the change of enthalpy in a chemical reaction is independent of the pathway between the initial and final states. This is useful for us in calculating change in enthalpy's because we can then use reactions with known enthalpy's and manipulate them to find the enthalpy of the d...
- Tue Jan 16, 2018 11:36 am
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Q 8.13
- Replies: 9
- Views: 993
Re: Q 8.13
No. In this question, it says that the cooling system that surrounds the cylinder absorbs 974 kJ of heat. If the system that surrounds the cylinder is absorbing heat, this means that the cylinder is releasing this heat and that is why it is negative. Heat absorbed by the cylinder would be positive, ...
- Tue Jan 16, 2018 11:30 am
- Forum: Phase Changes & Related Calculations
- Topic: 8.37
- Replies: 3
- Views: 245
Re: 8.37
For part a, you know that the vaporization of 0.579 moles of Ch4 requires 4.76 kJ of heat, so all you have to do is divide the amount of heat by the moles to find the enthalpy (kJ/mol). 4.76 kJ/0.579 mol=8.22 kJ/mol For part b it is very similar but instead of being given moles, you are given grams....
- Fri Dec 08, 2017 9:56 pm
- Forum: Bronsted Acids & Bases
- Topic: 12.17 How can you determine if an oxide is acidic, basic, or amphoteric
- Replies: 1
- Views: 755
Re: 12.17 How can you determine if an oxide is acidic, basic, or amphoteric
ad trouble with this question too. Reading page 468-469 in the textbook really helped me. It says at the end of section 12.3: metals form basic oxides, nonmetals form acidic oxides, and the elements on a diagonal line from beryllium to polonium and several d-block metals form amphoteric oxides. Ther...
- Mon Dec 04, 2017 11:35 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: 11.73 d
- Replies: 2
- Views: 387
Re: 11.73 d
Dr. Lavelle mentioned in lecture a quick way to do this problem. Since pressure is increasing and volume is decreasing. When that happens, you know that if there are more moles on the left, the reaction shifts right. If there are more moles on the right, the reaction shifts left. In this case there ...
- Mon Dec 04, 2017 11:30 pm
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: 12.17
- Replies: 3
- Views: 332
Re: 12.17
I had trouble with this question too. Reading page 468-469 in the textbook really helped me. It says at the end of section 12.3: metals form basic oxides, nonmetals form acidic oxides, and the elements on a diagonal line from beryllium to polonium and several d-block metals form amphoteric oxides. h...
- Mon Dec 04, 2017 11:27 pm
- Forum: Conjugate Acids & Bases
- Topic: HW 12.3
- Replies: 3
- Views: 626
Re: HW 12.3
I believe what you learned in high school was right. A conjugate base should only have the different of one hydrogen
- Mon Dec 04, 2017 11:25 pm
- Forum: Bronsted Acids & Bases
- Topic: 12.39
- Replies: 2
- Views: 422
12.39
Chapter 12 #39 asks: 12.39 Using data in Tables 12.1 and 12.2, place the following acids in order of increasing strength: HNO2, HClO2, NH3OH, (CH3)2NH2 Considering we won't have these tables to use during the exam, are we expected to be able to place acids in order of strength? If so, could someone ...
- Mon Nov 27, 2017 8:38 pm
- Forum: Hybridization
- Topic: Lone Pairs in Hybridization
- Replies: 4
- Views: 728
Lone Pairs in Hybridization
Can anyone explain how to write the hybridization of a compound with a lone pair? Does hybridization rely solely on the amount of regions of electron density or just the number of bonds excluding lone pairs?
thanks
thanks
- Sun Nov 26, 2017 11:48 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Post Assessment Chemical Equilibrium Part 2 #29 [ENDORSED]
- Replies: 6
- Views: 1054
Re: Post Assessment Chemical Equilibrium Part 2 #29 [ENDORSED]
oh I see. thank you!
- Sun Nov 26, 2017 3:47 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Post Assessment Equilibrium Part 2 #30
- Replies: 2
- Views: 2283
Re: Module Part 2 #30
Hey Sabrina, so when you convert everything to molarity values you get 0.167 M of Carbon and .05 M of H20. Those are the initial amount for of C and H20 in the ice table. Since Carbon is a solid it does not have an effect on the equilibrium values. We know the equilibrium concentration of H2 is 0.04...
- Sun Nov 26, 2017 3:27 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Post Assessment Chemical Equilibrium Part 2 #29 [ENDORSED]
- Replies: 6
- Views: 1054
Re: Post Assessment Chemical Equilibrium Part 2 #29 [ENDORSED]
how did you solve for x without knowing the value of k?
- Fri Nov 24, 2017 10:00 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Post Assessment Chemical Equilibrium Part 2 #29 [ENDORSED]
- Replies: 6
- Views: 1054
Post Assessment Chemical Equilibrium Part 2 #29 [ENDORSED]
Can anyone explain #29 from the Chemical Equilibrium Part 2 Post Assessment? Here it is 29. A researcher fills a 1.00 L reaction vessel with 1.84 x 10-4 mol of BrCl gas and heats it to 500 K. At equilibrium, only 18.3 % of the BrCl gas remains. Calculate the equilibrium constant, assuming the follow...
- Sun Nov 19, 2017 7:22 pm
- Forum: Hybridization
- Topic: Electron density question
- Replies: 4
- Views: 751
Re: Electron density question
yes each lone pair counts as another region of electron density
- Sun Nov 19, 2017 5:37 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: HW Question 4.27
- Replies: 4
- Views: 474
Re: HW Question 4.27
By replacing the -CH- in Benzene with a Nitrogen atom the dipole moments no longer cancel so it becomes a polar molecule.
- Wed Nov 08, 2017 9:53 am
- Forum: Lewis Structures
- Topic: 3.67B and Radicals
- Replies: 2
- Views: 414
Re: 3.67B and Radicals
Yes, this molecule has an odd number of electrons(19). So the two oxygens would both have octets. But there are not enough electrons for chlorine to have an octet. Chlorine then has one lone pair, and one single electron.
- Wed Nov 08, 2017 9:50 am
- Forum: Electronegativity
- Topic: 3.79
- Replies: 3
- Views: 1559
Re: 3.79
Also remember that atoms with a similar number of valence electrons are more likely to equally share electrons and have a more covalent character. This is in contrast to ionic character, where one atom is monopolizing the electrons.
- Sun Nov 05, 2017 7:28 pm
- Forum: Formal Charge and Oxidation Numbers
- Topic: 3.49
- Replies: 3
- Views: 633
3.49
Can anyone explain 3.49? I had the correct lewis structure but am not getting the formal charge correct. For part A, I had Formal Charge of Oxygen= 6 valence e - (1 lone pair +6/2) which gives a +2 charge, and the book says Oxygen has a +1 charge. For Nitrogen I had Formal Charge of Nitrogen=5 valen...
- Sun Nov 05, 2017 9:29 am
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: 3.9
- Replies: 2
- Views: 325
3.9
Can anyone explain 3.9 from the homework problems? I am confused on what the question is asking for. Also for the given electron configurations
a.) [Ar] 3d7
b.[Ar]3d6
are these possible? Doesn't the 4s orbital need to be filled for to have an electron in 3d?
thanks
a.) [Ar] 3d7
b.[Ar]3d6
are these possible? Doesn't the 4s orbital need to be filled for to have an electron in 3d?
thanks
- Tue Oct 24, 2017 10:22 pm
- Forum: Trends in The Periodic Table
- Topic: 2.59
- Replies: 1
- Views: 187
2.59
Can anyone explain #59 from Chapter 2? Aren't the ions isoelectronic?
thanks!
thanks!
- Sun Oct 22, 2017 10:18 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: 2. 39 [ENDORSED]
- Replies: 5
- Views: 751
Re: 2. 39 [ENDORSED]
I had the same question....so you can determine wether the electron is in the excited state or not by seeing if it correctly spaced in the boxes?
- Sun Oct 22, 2017 6:23 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: XYZ [ENDORSED]
- Replies: 4
- Views: 2192
XYZ [ENDORSED]
Hi!
I did electron configuration in high school using the 2p61, 2p62, 2p63 model... can someone please explain the x,y,z we are using in this class more clearly? thanks!
I did electron configuration in high school using the 2p61, 2p62, 2p63 model... can someone please explain the x,y,z we are using in this class more clearly? thanks!
- Fri Oct 13, 2017 9:50 pm
- Forum: Photoelectric Effect
- Topic: Photoelectric Effect Post-Module Assessment 32B, 33 [ENDORSED]
- Replies: 3
- Views: 1353
Re: Photoelectric Effect Post-Module Assessment 32B, 33 [ENDORSED]
For 32B you know Kinetic energy of the electron and you know the work function so you are able to solve for the energy of the photon. KE=4.200 x10^-19 J Work function=3.607 x 10^-19 J, which you know from the previous question because there was zero kinetic energy. So then you do: Energy(photon)=Kin...
- Fri Oct 13, 2017 9:40 pm
- Forum: Photoelectric Effect
- Topic: Question 34 from Photoelectric Effect Post Module Assessment
- Replies: 1
- Views: 354
Question 34 from Photoelectric Effect Post Module Assessment
Can anyone help me with question 34 from this post assessment? I said C for #33 which was correct but couldn't figure out #34. Thanks! 33. Molybdenum metal must absorb radiation with a minimum frequency of 1.09 x 1015 s-1 before it can emit an electron from its surface. Answer the following two ques...
- Thu Oct 05, 2017 5:21 pm
- Forum: Molarity, Solutions, Dilutions
- Topic: E9 [ENDORSED]
- Replies: 5
- Views: 4782
Re: E9 [ENDORSED]
dividing by 246.48 g.mol^-1 gives the molar mass of the entire thing, and it just asks for oxygen so you multiply by 11 because there are 11 moles of Oxygen in the equation. Then after you convert to atoms
- Thu Oct 05, 2017 5:07 pm
- Forum: Limiting Reactant Calculations
- Topic: M11
- Replies: 5
- Views: 5096
Re: M11
I have the same question.Couldn't quite follow the solutions manual