Search found 144 matches
- Sat Mar 15, 2014 9:54 pm
- Forum: *Organic Reaction Mechanisms in General
- Topic: 2012 final Q5
- Replies: 1
- Views: 542
Re: 2012 final Q5
the O with the double bond becomes a single bond as those electrons are pointed to the oxygen atom* see the other arrow curving to the right. When that occurs, the carbon that is connected to the oxygen is a carbocation and has a positive charge.
- Fri Mar 14, 2014 5:58 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 2009 Final pg 206 Q 4E
- Replies: 1
- Views: 397
Re: 2009 Final pg 206 Q 4E
The equation when solving for H is essentially E x 5 /.0592 = log Q Solve for Q = (10^E*5 /.0592) And solve for H. Take the negative log and you should get -1.667 OR When you have log Q, plug in the other concentrations and you should get log( 1000*[H+]^8) and separate the inside by the addition of ...
- Thu Mar 13, 2014 4:33 pm
- Forum: *Cycloalkenes
- Topic: Final 2013 6a Numbering a hexene
- Replies: 8
- Views: 2326
Re: Final 2013 6a
I believe there is a rule that if there is a substituent on one of the double bonds, then that carbon is labeled as the first one.
- Thu Mar 13, 2014 4:04 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 2012 final Q3 (a)
- Replies: 3
- Views: 665
Re: 2012 final Q3 (a)
That's how I determined the anode. Whichever reduction potentiometer was more negative/least positive!
- Thu Mar 13, 2014 3:58 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Graphing
- Replies: 1
- Views: 611
Re: Graphing
The units should be in Kelvin because the formula k=e^-Ea/RT
The constant R is in J/mol*K
The constant R is in J/mol*K
- Thu Mar 13, 2014 3:57 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: Quiz 1 Problem
- Replies: 2
- Views: 750
- Thu Mar 13, 2014 3:26 pm
- Forum: General Rate Laws
- Topic: Reaction Rates
- Replies: 2
- Views: 538
Re: Reaction Rates
The reaction rate is decreased by 1/4 because Cl2 is second order. (.5x)^2 = .25(x^2)
- Thu Mar 13, 2014 1:35 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: 7.19 Book Question
- Replies: 2
- Views: 509
Re: 7.19 Book Question
The heat given off by the copper is gained by the surroundings: water. qsystem= -qsurr
So use the formula q= mCdeltaT and solve for the final temp
You need to get the specific heat capacities of copper and water from a constant sheet.
So use the formula q= mCdeltaT and solve for the final temp
You need to get the specific heat capacities of copper and water from a constant sheet.
- Thu Mar 13, 2014 11:24 am
- Forum: *Alkanes
- Topic: Numbering based on alphabetical or substituent size?
- Replies: 5
- Views: 899
Re: Numbering based on alphabetical or substituent size?
Just to confirm, when you have sec or ter before an alkyl, do you disregard those italicized prefixes when alphabetizing. But when you have iso or neo, those are accounted for in the alphabetizing
- Thu Mar 13, 2014 2:07 am
- Forum: *Alkenes
- Topic: Relative Stabilities of Alkenes/Alkynes
- Replies: 3
- Views: 2451
Re: Relative Stabilities of Alkenes/Alkynes
Internal alkenes/alkynes are more stable than terminal ones because when the bond is internal and connected to more than one carbon-secondary, tertiary, quaternary--, the pi bonds are more stabilized by the surrounding carbons. Also, I think that having a terminal double/triple bond would allow more...
- Sun Mar 09, 2014 5:52 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Residual entropy
- Replies: 2
- Views: 707
Re: Residual entropy
NO also has resonances; double and triple bonds
- Sun Mar 09, 2014 5:49 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Heat capacity
- Replies: 2
- Views: 607
Re: Heat capacity
Ethane has more degrees of freedom than that of ethene. Ethene has a double bond so it stays in a rigid structure. Each degree of freedom contributes 1/2R to the heat capacity Cv,m
- Sat Mar 08, 2014 3:01 pm
- Forum: *Constitutional and Geometric Isomers (cis, Z and trans, E)
- Topic: Final 2011 Q7B
- Replies: 3
- Views: 717
Final 2011 Q7B
Question 7B on the Winter 2011 Final Exam states: "Give the full IUPAC name for this molecule." The molecule features a (Z) alkene with a carboxylic acid, chlorine, and alkyl group as its substituents. The molecule is labeled as cis-Z and the sustituents are H, C2H5 on the left and Cl, and...
- Wed Mar 05, 2014 10:29 am
- Forum: Administrative Questions and Class Announcements
- Topic: *Additional Forum Features*
- Replies: 25
- Views: 4224
Re: *Additional Forum Features*
It'd be cool if we could "like" or mark a post as "helpful."
- Tue Mar 04, 2014 4:07 pm
- Forum: *Alkenes
- Topic: Naming alkene along with halogen
- Replies: 2
- Views: 552
Re: Naming alkene along with halogen
On page 49, the 3 does not indicate where the double bond is; it is indicating the -iodo part of the hydrocarbon.
Double bonds have priority when numbering, so in 3-iodobutene, you number the carbons from the left to right.
Double bonds have priority when numbering, so in 3-iodobutene, you number the carbons from the left to right.
Re: naming
Edit: in the hw problems and the practice finals, it seems both are accepted unless specified.
- Tue Mar 04, 2014 3:33 pm
- Forum: *Alkanes and Substituted Alkanes (Staggered, Eclipsed, Gauche, Anti, Newman Projections)
- Topic: The Correct Newman Projection
- Replies: 2
- Views: 8449
Re: The Correct Newman Projection
"any correct newman projection" refers to eclipsed, gauche, and anti. +2
to get the other 2 points, "THE correct" one is the anti conformation which is the most stable.
to get the other 2 points, "THE correct" one is the anti conformation which is the most stable.
- Mon Mar 03, 2014 10:48 pm
- Forum: *Chem3D
- Topic: Why do bond angle values change?
- Replies: 4
- Views: 2363
Re: Why do bond angle values change?
I googled energy minimization and several websites state: "Perform an energy minimization of the system, by iteratively adjusting atom coordinates. Iterations are terminated when one of the stopping criteria is satisfied. At that point the configuration will hopefully be in local potential ener...
Re: 33
The OH- and Br- are substituted. The first step involves the nucleophile OH- attaching to the CH3Br- to reach the transition state. The reaction shown to be bimolecular because the rate limiting step has an overall order of 2
Page 159 of the Intro to Organic Chemistry text.
Page 159 of the Intro to Organic Chemistry text.
- Thu Feb 27, 2014 11:27 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: to determine the overall reaction
- Replies: 6
- Views: 1986
Re: to determine the overall reaction
OH- is left in the rate law because it's concentration is so small (from the fact that [H3O+]=[OH-]= 10^-7) that it does have influence.
- Tue Feb 25, 2014 11:00 pm
- Forum: *Enzyme Kinetics
- Topic: Rate constant
- Replies: 2
- Views: 625
Re: Rate constant
k= Ae^-Ea/RT
When an enzyme is added, it lowers the activation energy barrier Ea. Therefore, the rate constant will increase.
When an enzyme is added, it lowers the activation energy barrier Ea. Therefore, the rate constant will increase.
- Tue Feb 25, 2014 10:47 pm
- Forum: First Order Reactions
- Topic: Quiz 2 2014 Prep: Question 4
- Replies: 2
- Views: 777
Re: Quiz 2 2014 Prep: Question 4
You're probably referring to the question: Calculate the time required for the concentration of N2O... If you notice the rate constant has the units 1/s, that tells you that it's a first order reaction. First order reactions have the units 1/s because… rate= k[A] Units: M/s= k* M k= 1/s So use the i...
- Tue Feb 25, 2014 10:43 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: rate constant
- Replies: 1
- Views: 565
Re: rate constant
k= Ae^-Ea/RT
Here the rate constant changes as a function of temperature. Ea is also a constant*
Here the rate constant changes as a function of temperature. Ea is also a constant*
- Tue Feb 25, 2014 11:01 am
- Forum: General Rate Laws
- Topic: Rate law with intermediates
- Replies: 3
- Views: 890
Re: Rate law with intermediates
Thank you, you're right. One method is pre equilibrium and the other is steady state approx. Just recalled that from lecture...
- Sat Feb 22, 2014 7:39 pm
- Forum: First Order Reactions
- Topic: Question 14.23 Integrated Rate Law
- Replies: 2
- Views: 540
Re: Question 14.23 Integrated Rate Law
Page 73 of the course reader will answer your question. It's because in a real life instance, the balanced reaction is unknown.
when you collect data, plot the points and obtain at straight line, then you can get the k value (proportionality constant)
when you collect data, plot the points and obtain at straight line, then you can get the k value (proportionality constant)
- Sat Feb 22, 2014 2:08 pm
- Forum: General Rate Laws
- Topic: Rate law with intermediates
- Replies: 3
- Views: 890
Rate law with intermediates
Step 1 A fast bimolecular dimerization and its reverse: NO + NO --> N2O2 Rate of formation of N2O2 = k1[NO]2 N2O2 ---> NO + NO Rate of formation of N2O2 = k1'[N2O2] Step 2 A slow bimolecular reaction in which an O2 molecule collides with the dimer: O2 + N2O2 ---> NO2 + NO2 Rate of consumption of N2...
- Sat Feb 22, 2014 1:12 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: to determine the overall reaction
- Replies: 6
- Views: 1986
Re: to determine the overall reaction
In dilute solutions where H2O is the solvent, it can be omitted because the concentration does not noticeably change.
- Sat Feb 22, 2014 12:15 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: to determine the overall reaction
- Replies: 6
- Views: 1986
Re: to determine the overall reaction
I just finished this problem too… Yes, the overall reaction is determined that way by adding all three reactions and canceling what's on both sides. The rate of reaction is determined by the slowest step, which in this case is the 2nd step (very slow). Intermediates cannot be in the rate law, so you...
- Mon Feb 10, 2014 6:32 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: Midterm Winter 2013 Q5B
- Replies: 1
- Views: 500
Re: Midterm Winter 2013 Q5B
All the central atoms are carbon surrounded by 4 molecules. The difference between these four species though is the size. Larger size, complex molecules have higher entropies than smaller size molecules because they have higher vibrational energies (energy levels are closer together when larger)
- Mon Feb 10, 2014 6:27 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: 2008 Midterm Q6A
- Replies: 1
- Views: 361
Re: 2008 Midterm Q6A
To determine the charge of the Mn, you look at the molecule overall and its oxidation numbers. So permanganate, MnO4- has an overall charge of -1. Oxygen molecules have an oxidation number of -2 normally and there are 4 of them so -2 x 4 = -8. So now you have to determine what value + -8 = -1 Solvin...
- Mon Feb 10, 2014 6:23 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 2008 5A
- Replies: 1
- Views: 267
Re: 2008 5A
The E value of the reduction potential that is the more negative/least positive has a higher reducing power , or is more likely to be oxidized ; therefore, at the anode, Cd is oxidized to Cd2+. To calculate the standard E of the cell, you can either flip the sign of the reduction potential of the an...
- Sat Feb 08, 2014 5:32 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Winter 2009 midterm Question 1
- Replies: 1
- Views: 372
Re: Winter 2009 midterm Question 1
You're right. The combustion of glucose is exothermic and has a negative delta H value. The heat released from the reaction is now used to heat the water in the child's body, so the value changes to positive value. the heat lost by the system = the heat gained by the surroundings q sys = -q surr In ...
- Thu Feb 06, 2014 12:05 am
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Calculating Q, Ch 12 problem 41
- Replies: 2
- Views: 1263
Re: Calculating Q, Ch 12 problem 41
Chapter 13 problem 41 What you have to do is write out the half reactions for the cathode and the anode. When you add the two equations together, you will have concentrations and partial pressures on the product side and the same for the reactants. Then you can write Q by taking the multiplication ...
- Wed Feb 05, 2014 11:55 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: 13.103
- Replies: 2
- Views: 555
Re: 13.103
That was very helpful! I appreciate your comprehensive explanation.
- Wed Feb 05, 2014 4:20 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: 13.103
- Replies: 2
- Views: 555
13.103
Use the data in Appendix 2B and the fact that, for the half- reaction F2(g) + 2 H+(aq) + 2 e– --> 2 HF(aq), E° = +3.03 V, to calculate the value of Ka for HF. How are the significant figures applied in this problem? The book uses natural log, so when solving lnK = (2)(-.16)/(.02569)= -12.45 …. the ...
- Tue Feb 04, 2014 5:45 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: 13.41 Reaction quotients
- Replies: 2
- Views: 420
Re: 13.41 Reaction quotients
Thank you!
- Tue Feb 04, 2014 2:13 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: 13.41 Reaction quotients
- Replies: 2
- Views: 420
13.41 Reaction quotients
(c)Pt(s)|Cl2(g, 250 Torr)|HCl(aq, 1.0 M)||HCl(aq, 0.85 M)|H2(g, 125 Torr)|Pt(s) I know how to solve the problem using the correct equations; I just have a question on the reaction quotients. The reaction quotients have both partial pressure and concentrations, which i have never seen before. Is this...
- Fri Jan 31, 2014 4:33 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 13.11
- Replies: 4
- Views: 629
Re: 13.11
Actually, I think I figured it out. I set up O2 --> H+ and O2 ---> OH- and balanced the half reactions by using the procedures if in acidic solution and basic solution, respectively. Probably had to do with the fact that water is amphoteric and has acidic and basic properties?
- Fri Jan 31, 2014 3:30 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 13.11
- Replies: 4
- Views: 629
13.11
Write the half-reactions and the balanced equation for the cell reaction for each of the following galvanic cells:
(d) Pt(s)|O2(g)|H+(aq)||OH-(aq)|O2(g)|Pt(s)
How do I set up the two half reactions for each one? This one seems different from all the others.
(d) Pt(s)|O2(g)|H+(aq)||OH-(aq)|O2(g)|Pt(s)
How do I set up the two half reactions for each one? This one seems different from all the others.
- Mon Jan 27, 2014 7:49 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 10.109
- Replies: 1
- Views: 1496
10.109
(a) Calculate K at 25°C for the reaction Br2(g) #4 2 Br(g) from the thermodynamic data provided in Appendix 2A. (b) What is the vapor pressure of liquid bromine? (c) What is the partial pressure of Br(g) above the liquid in a bottle of bromine at 25°C? I only have a question for part c. The solutio...
- Sat Jan 25, 2014 9:13 pm
- Forum: Phase Changes & Related Calculations
- Topic: Phase Change/heat curve Graphs
- Replies: 2
- Views: 581
- Sat Jan 25, 2014 12:19 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Do all pressure units operate in the same gradient?
- Replies: 2
- Views: 376
Re: Do all pressure units operate in the same gradient?
I would not think so because a quotient is just a ratio and if you convert the units to another type of units, it still maintains that same ratio. For ex, 1 kPa to 2 kPa is still the same ratio in atm--1/2.
- Sat Jan 25, 2014 12:14 pm
- Forum: Phase Changes & Related Calculations
- Topic: Phase Change/heat curve Graphs
- Replies: 2
- Views: 581
Phase Change/heat curve Graphs
When analyzing a heat curve graph (heat vs Temp), I noticed that the slopes matter when heating a solid, liquid, and vapor. So if you're trying to determine the molar specific heat, C which is equal to q /[(moles) X deltaT], then you would just look at the q/ delta T (moles is just a constant)…. The...
- Sat Jan 25, 2014 11:53 am
- Forum: Administrative Questions and Class Announcements
- Topic: Quiz 1
- Replies: 3
- Views: 659
Quiz 1
For Quiz one, there are a couple problems I have encountered that require certain values (specific heat of iron (Q2) and heat of formation of glucose(Q5)) that are not provided in the constant sheets. These, of course, will be given to us on the quiz, correct?
Thank you
Thank you
- Fri Jan 24, 2014 9:18 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: HMW: 8.59 (a) and (c)
- Replies: 3
- Views: 678
Re: HMW: 8.59 (a) and (c)
Delta G is calculated by the (sum of Gf of the products - sum of Gf of reactants).
O2 was not included because that value is 0.
:)
O2 was not included because that value is 0.
:)
- Fri Jan 24, 2014 6:19 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 8.103
- Replies: 4
- Views: 776
Re: 8.103
but for the three reactions of the three isomers, the solution manual has set up equations for the following and found the del S to be 1) isomer 1--> isomer 2 del S= -4.33 J/mol-K 2) isomer 1-->isomer 3 del S = -7.11 3) isomer 2-->isomer 3 del S= -2.78 How do you find the Sm for each isomer. my atte...
- Thu Jan 23, 2014 12:30 pm
- Forum: Phase Changes & Related Calculations
- Topic: units
- Replies: 2
- Views: 419
Re: units
The entropy values are given in Joules in the appendix, so remember to divide entropy by 1000 to keep everything in kJ. Also for the sake of simplicity when using the equations such as delta G= delta H - TdeltaS
- Thu Jan 23, 2014 12:28 pm
- Forum: Phase Changes & Related Calculations
- Topic: trouton's rule
- Replies: 2
- Views: 582
Re: trouton's rule
I know that organic means containing carbon, C. So any liquid hydrocarbons, alcohols, etc. look at appendix 2A, A16
- Thu Jan 23, 2014 12:26 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: entropy and volume
- Replies: 7
- Views: 1031
Re: entropy and volume
Precisely!
dS = nR ln (P1/P2) --the formula when pressure is involved. notice that P initial is on top..
dS = nR ln (P1/P2) --the formula when pressure is involved. notice that P initial is on top..
- Wed Jan 22, 2014 10:04 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 8.103
- Replies: 4
- Views: 776
Re: 8.103
Thank you!
- Wed Jan 22, 2014 7:00 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 8.103
- Replies: 4
- Views: 776
8.103
Three isomeric alkenes have the formula C4H8 (see the following table). (a) Draw Lewis structures of these compounds. (b) Calculate dG°, dH°, and dS° for the three reactions that interconvert each pair of compounds. (c) Which isomer is the most stable? (d) Rank the isomers in order of decreasing Sm...
- Wed Jan 22, 2014 4:08 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: Self test 8.3A
- Replies: 1
- Views: 532
Re: Self test 8.3A
dS = R*ln (V2/V1)
If it is compress to 1/3 its initial volume, then V2= (1/3) V1
Plug in V2 into the equation and the V1 will cancel, so you have ln (1/3)
R=8.314 J/K*mol
change in molar entropy = 8.314 * ln (1/3) = -9.13 J/mol*K
If it is compress to 1/3 its initial volume, then V2= (1/3) V1
Plug in V2 into the equation and the V1 will cancel, so you have ln (1/3)
R=8.314 J/K*mol
change in molar entropy = 8.314 * ln (1/3) = -9.13 J/mol*K
- Tue Jan 21, 2014 11:32 am
- Forum: Calculating Work of Expansion
- Topic: Work of an isothermal reversible expansion
- Replies: 4
- Views: 728
Re: Work of an isothermal reversible expansion
The equation you have written is derived from integration, which takes all the small changes in volume that occur. Not just the initial and final states, I believe.
- Tue Jan 21, 2014 11:30 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 8.43
- Replies: 2
- Views: 869
Re: 8.43
Thanks Iris!
- Mon Jan 20, 2014 6:56 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Question 7.115- Chemistry Connections Problem
- Replies: 1
- Views: 857
Re: Question 7.115- Chemistry Connections Problem
They took the enthalpy of combustion values given in Appendix 2A and multiplied it by the stochiometric ratios particular to each combustion reaction to get kJ/ mol CO2.
- Mon Jan 20, 2014 6:53 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: entropy and volume
- Replies: 7
- Views: 1031
Re: entropy and volume
I would think so given from the formula dS = nR ln (V2/V1).
"n" and "R" would be given constants and analyzing the graph of a ln function, an increase in volume would make V2/V1 positive and as the final volume gets larger, the magnitude of "ln (V2/V1)" does as well.
"n" and "R" would be given constants and analyzing the graph of a ln function, an increase in volume would make V2/V1 positive and as the final volume gets larger, the magnitude of "ln (V2/V1)" does as well.
- Thu Jan 16, 2014 11:06 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 8.43
- Replies: 2
- Views: 869
8.43
Problem: Calculate the standard entropy of vaporization of water at 85°C, given that its standard entropy of vaporization at 100.°C is 109.0 J/K-mol and the molar heat capacities at constant pressure of liquid water and water vapor are 75.3 J/mol-k and 33.6 J/mol-k, respectively, in this range So t...
- Tue Jan 14, 2014 10:06 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: HW 7.67 Calculating the reaction enthalpy
- Replies: 1
- Views: 349
Re: HW 7.67 Calculating the reaction enthalpy
NH (g) + HCl(g) --> NH4Cl(s) N2(g) + 3 H2(g) ---> 2 NH3(g) N2(g) + 4 H2(g) + Cl2(g) ---> 2 NH4Cl(s) You want: H2 + Cl2 --> 2HCl Do not divide the second equation by 3, reverse it!Since you reverse the 1st equation the NH3 is on the product side and if you reverse the second equation the NH3 will be...
- Sun Jan 12, 2014 12:27 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Hmw: 7.53
- Replies: 5
- Views: 2701
Re: Hmw: 7.53
Sorry I explained the whole problem; I interpreted your question incorrectly :(
- Sun Jan 12, 2014 12:25 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: 7.29 C
- Replies: 3
- Views: 472
Re: 7.29 C
Under the enthalpy section of the textbook, there's a chart that tells you what molecular geometry corresponds to the value of Cvm.
- Sun Jan 12, 2014 12:14 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Hmw: 7.53
- Replies: 5
- Views: 2701
Re: Hmw: 7.53
Treat them as separate ions. So Zn2+ and 2 Cl-
- Sun Jan 12, 2014 12:04 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Work = -P X delta-V
- Replies: 4
- Views: 2309
Re: Work = -P X delta-V
When you multiple the variables, you get the units: L-atm. however work is the units Joules because it's a measure of energy. They multiply by this constant provided in the constant sheet/book. which is 101.325 J = 1 L-atm
- Sun Jan 12, 2014 12:02 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 7.71 [ENDORSED]
- Replies: 5
- Views: 2888
Re: 7.71 [ENDORSED]
If you mean delta H rxn (overall heat of the reaction), you would multiply the coefficients of the first equation by 2. By doing so, you must also multiply the delta H of that reaction. When you add the two equations together, the NO2 cancels out on both sides. Also add the heats of the reactions to...
- Sun Jan 12, 2014 11:52 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: calculating net formation of gas (moles)
- Replies: 4
- Views: 754
Re: calculating net formation of gas (moles)
delta H = delta U + (change in moles) R*T
-3920 kJ = delta U + (-3) (8.314) (298K)
*Make sure to convert [(-3) (8.314) (298K)] to kJ because the units are in Joules.
-3920kJ + (3)(8.314)(298) / 1000 = delta U
-3912.567
3 sig figs from initial "-3920"
so -3.91 x 10^3 kJ
-3920 kJ = delta U + (-3) (8.314) (298K)
*Make sure to convert [(-3) (8.314) (298K)] to kJ because the units are in Joules.
-3920kJ + (3)(8.314)(298) / 1000 = delta U
-3912.567
3 sig figs from initial "-3920"
so -3.91 x 10^3 kJ
- Sun Jan 12, 2014 11:26 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Hmw: 7.53
- Replies: 5
- Views: 2701
Re: Hmw: 7.53
7.53 Hydrochloric acid oxidizes zinc metal in a reaction that produces hydrogen gas and chloride ions. A piece of zinc metal of mass 8.5 g is dropped into an apparatus containing 800.0 mL of 0.500 M HCl(aq). If the initial temperature of the hydrochloric acid solution is 25°C, what is the final tem...
- Wed Jan 08, 2014 6:44 pm
- Forum: Phase Changes & Related Calculations
- Topic: 7.17b
- Replies: 2
- Views: 530
Re: 7.17b
It's amount of heat required to raise the temperature of the water from 22- 100 degrees celsius. The calculation is mass of the water (400g) x specific heat capacity of water (4.184) x Change in temperature (373-295K) which equals to 1.3 x 10^5.
- Sat Dec 07, 2013 10:12 pm
- Forum: *Indicators
- Topic: What do you base the indicator on?
- Replies: 1
- Views: 857
Re: What do you base the indicator on?
At the stochiometric point because pKin = pKa. pKa= pH @ stochiometric point in weak-acid strong base weak base- strong acid titrations
- Sat Dec 07, 2013 9:44 pm
- Forum: Empirical & Molecular Formulas
- Topic: molecular formulas
- Replies: 1
- Views: 416
Re: molecular formulas
empirical formula is the smallest whole number ratio of atoms; therefore, the molecular formula cannot be smaller than the empirical. However, the molecular formula and empirical formula can be the same in some cases when the ratio of the molecular formula cannot be reduced.
- Sat Dec 07, 2013 6:55 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Unpaired Electrons
- Replies: 1
- Views: 5079
Re: Unpaired Electrons
Look valence electron configuration and look at the very last subshell (3s, 2p, etc.). If it's p 2 then there are 2 unpaired electrons. Why? The p subshell has 3 orbitals px, py, pz. Each can hold a max of 2 electrons (pauli exclusion principle). Then, you have 1 e- in px and 1e- in py (following Hu...
- Sat Dec 07, 2013 6:41 pm
- Forum: Molarity, Solutions, Dilutions
- Topic: G. 25
- Replies: 1
- Views: 1261
Re: G. 25
First, you would set up the dilution equation M 1 V 1 = M 2 V 2 because you're using concentrated acid and diluting it to make a solution with a certain molarity. To find the volume you would need, divide the M 1 by both sides so you have V 1 =… M 2 V 2 = the number of moles HCl that you want, which...
- Thu Dec 05, 2013 9:50 pm
- Forum: *Titrations & Titration Calculations
- Topic: 12.33
- Replies: 2
- Views: 470
Re: 12.33
If you think about it on a step by step basis, from the beginning to the stochiometric pt., the base is neutralizing the acid by taking a proton and creating more conjugate base. At the stochiometric point, the all acetic acid has been neutralized into its conjugate base. You found the volume of nee...
- Thu Dec 05, 2013 8:48 pm
- Forum: *Titrations & Titration Calculations
- Topic: Different equations
- Replies: 1
- Views: 325
Re: Different equations
When you have a weak acid and a strong base titration and you titrate with a strong base before reaching the stochiometric point, you use the equation with the weak acid giving off a proton to water, making hydronium and the conjugate base. That's because the base is still pulling protons off the we...
- Wed Dec 04, 2013 6:08 pm
- Forum: Identifying Acidic & Basic Salts
- Topic: 11.77
- Replies: 1
- Views: 418
Re: 11.77
CH3NH3+ would act as an acid because if you know that it's the conjugate acid of the weak base CH3NH2. (pg 451)
conjugate acids of weak bases give off protons...
conjugate acids of weak bases give off protons...
- Wed Dec 04, 2013 6:07 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: Writing out reactions between acids and bases
- Replies: 2
- Views: 388
Re: Writing out reactions between acids and bases
NaCl exists as ions in aqueous solutions, not as a solid; therefore it is written as NaCl (aq). Mg(OH)2 is a solid because according to the solubility rules, compounds with OH- are insoluble except if they have cations of Ca2+, Ba2+, Sr2+. We didn't learn solubility rules in class but if you've take...
- Wed Dec 04, 2013 5:58 pm
- Forum: Bronsted Acids & Bases
- Topic: How strong of an acid is HPO4 -?
- Replies: 3
- Views: 1167
Re: How strong of an acid is HPO4 -?
Probably very weak because if you look at the acid H3PO4, which has 3 ionization constants, the one for HPO4- (Ka3) is 2.1 x 10-13. I was just looking at the chart on page 455 for polyprotic acids
- Tue Dec 03, 2013 3:57 pm
- Forum: Bronsted Acids & Bases
- Topic: 11.19
- Replies: 3
- Views: 599
Re: 11.19
Metal oxides tend to be basic whereas nonmetal oxides turn acidic.
Example: nonmetal oxide co2 + h20 > h2co3 ( acidic)
Metal oxide CaO + h2o > Ca(OH)2
Oxides with metalloid are amphoteric.
Example: nonmetal oxide co2 + h20 > h2co3 ( acidic)
Metal oxide CaO + h2o > Ca(OH)2
Oxides with metalloid are amphoteric.
- Mon Dec 02, 2013 10:08 pm
- Forum: *Titrations & Titration Calculations
- Topic: Reversal on Titration Curve
- Replies: 2
- Views: 501
Re: Reversal on Titration Curve
Just look at the graph sideways. However you would never draw a graph like that because the x axis is the independent variable and the y axis is the dependent variable. The volume of titrant added is the independent variable (variable being manipulated) and the pH is "dependent" on that.
- Mon Dec 02, 2013 10:06 pm
- Forum: Bronsted Acids & Bases
- Topic: NH3+ + HCl ---> NH4Cl
- Replies: 4
- Views: 10166
Re: NH3+ + HCl ---> NH4Cl
I would say NH3 is a Bronsted base because it accepts a proton. and ammonium chloride is separate in an aqueous solution.
To your second question: HCl is a bronsted acid (proton donor). Lewis acids accept electron pairs, which it does not in aqueous solutions
To your second question: HCl is a bronsted acid (proton donor). Lewis acids accept electron pairs, which it does not in aqueous solutions
- Mon Dec 02, 2013 8:55 pm
- Forum: Polyprotic Acids & Bases
- Topic: Determining what is a polyprotic acid/base
- Replies: 5
- Views: 989
Re: Determining what is a polyprotic acid/base
Well what if you have a salt in solution, CH3NH3+ and Cl-. The cation that acts as an acid donates one H and is not polypro tic (at least i think so). Also NH4+. It donates 1 proton, although having 4 multiple hydroden atoms.
- Mon Dec 02, 2013 8:49 pm
- Forum: *Making Buffers & Calculating Buffer pH (Henderson-Hasselbalch Equation)
- Topic: 11.21a
- Replies: 4
- Views: 488
Re: 11.21a
You can find the polyprotic Ka values on page 455 of the textbook. Good idea to bookmark them because the problems refer to tables 11.1,2 etc a lot.
- Mon Dec 02, 2013 12:44 pm
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: pKa and its relationship with how acidic something is
- Replies: 3
- Views: 33505
Re: pKa and its relationship with how acidic something is
If you have a low pKa, that means that your Ka value is high. Give by the equation -log [Ka] = pKa -> 10^-pKa = Ka. A lower pKa means the Ka value is higher and a higher Ka value means the acid dissociates more readily because it has a larger concentration of Hydronium ions (H 3 O + ). If you have a...
- Mon Dec 02, 2013 12:05 pm
- Forum: Lewis Acids & Bases
- Topic: HCl
- Replies: 4
- Views: 787
Re: HCl
The species HCl is a Lewis acid because it acts as a proton donor, the proton being h+
- Sat Nov 30, 2013 6:28 pm
- Forum: Conjugate Acids & Bases
- Topic: 2008 final exam 6.c
- Replies: 2
- Views: 407
Re: 2008 final exam 6.c
I did that problem earlier. The answer was that [H30+] does not equal [HA]. Since it's a weak acid, not all of the H+ is released from the acid (deprotonated) to make H30+. pKa of 10 means the Ka is very small, also reinforcing that the acid doesn't deprotonate readily like a strong acid.
- Wed Nov 27, 2013 11:54 pm
- Forum: Bronsted Acids & Bases
- Topic: 11.71
- Replies: 2
- Views: 374
Re: 11.71
When in aqueous solution, the ions exist as nh4+ and br-. Br- is the conjugate base of a strong acid therefore it is neutral according to the rules in the text and in course reader I believe, and has no effect on the solution. Nh4+ is the conjugate acid of a weak base nh3. The rule is: conjugate aci...
- Wed Nov 27, 2013 11:49 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: 11.75c
- Replies: 4
- Views: 511
Re: 11.75c
I think it involves the concept that small, highly charged cations can be acidic... They are surrounded by water molecules (hydration shell) and can pull the electrons of the oxygen on one H2o closely and kick out a hydrogen, resulting in an oh and h+. The equation is similar to 11.71(e) if you look...
- Wed Nov 27, 2013 11:42 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: 11.37, part E
- Replies: 3
- Views: 542
Re: 11.37, part E
The solution from the back of the textbook is in the correct order. Is says the h2SeO3 is the weakest acid.
- Tue Nov 26, 2013 11:28 am
- Forum: Identifying Acidic & Basic Salts
- Topic: 11.19
- Replies: 8
- Views: 1208
Re: 11.19
Metalloids are amphoteric
- Thu Nov 21, 2013 12:55 pm
- Forum: Identifying Acidic & Basic Salts
- Topic: 11.19
- Replies: 8
- Views: 1208
Re: 11.19
Nonmetal oxides tend to be acidic when reacted with water and metal oxides tend to be basic when reacted with water. The reason is that oxides O2- and water form hydroxide ions
- Wed Nov 20, 2013 9:58 pm
- Forum: Naming
- Topic: Copper vs. Cupperate
- Replies: 2
- Views: 933
Re: Copper vs. Cupperate
Its actually Cuprate, using the latin prefix.
Use Cuprate when the overall charge in the complex is negative. Use copper if it is neutral or positive.
Use Cuprate when the overall charge in the complex is negative. Use copper if it is neutral or positive.
- Tue Nov 19, 2013 7:28 pm
- Forum: Naming
- Topic: Am I alphabetizing Wrong
- Replies: 5
- Views: 917
Re: Am I alphabetizing Wrong
The old convention was to alphabetize but put the anions before the neutral ligands. That's why OH was before NH3. I think the new convention is to simply alphabetize the actually symbols in the formula.
- Tue Nov 19, 2013 6:59 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Equilibrium calculations in the ICE Table
- Replies: 3
- Views: 606
Re: Equilibrium calculations in the ICE Table
Do you mean converting concentration (mol/L) using the stochiometric ratios. And I believe you can use that on the C row, where you usually put -x, +x, etc.,
- Thu Nov 14, 2013 8:50 pm
- Forum: Naming
- Topic: Order of Coordination Compound inside the brackets.
- Replies: 2
- Views: 502
Re: Order of Coordination Compound inside the brackets.
For your first question, you alphabetize. I2, nh3, then o2. Same rule applies to multiple anions.
- Thu Nov 14, 2013 8:42 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: 16.31
- Replies: 1
- Views: 459
Re: 16.31
Oxygen is linear and can only bring to the metal cation on one side/ it can coordinate to one bonding region and contribute a pair of electrons. Although it has two paid of electrons. For other molecules with complex geometries that can surround the action like a claw, those ligand s can occupy mult...
- Thu Nov 14, 2013 8:35 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Cubic equations for x
- Replies: 2
- Views: 489
Re: Cubic equations for x
We won't be dealing with cubics, but if you do run into them in the future, you could graph it and find the zeros (on a graphing calc of course)
- Thu Nov 14, 2013 8:31 pm
- Forum: Naming
- Topic: Cyanido vs. Isocyanido
- Replies: 3
- Views: 898
Re: Cyanido vs. Isocyanido
The one that is linked to the metal ion will be underlined.
- Thu Nov 14, 2013 8:30 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Homework 10.27
- Replies: 2
- Views: 461
Re: Homework 10.27
I think you got mixed up with the h and hcl reaction which is 4 x 10^-18.
The K value for the reaction according to the chart says 160. It's the third reaction listed :)
The K value for the reaction according to the chart says 160. It's the third reaction listed :)
- Thu Nov 14, 2013 8:22 pm
- Forum: Naming
- Topic: Problem from textbook ch.16, 16.29 d
- Replies: 3
- Views: 632
Re: Problem from textbook ch.16, 16.29 d
The book says prefixes aren't included when you alphabetize. But for polydentates and Ligands with a prefix already in the name, such as ethylene diaamine, you use the prefixes bis, tris, and tetrakis. Oxalate is a bidentate; therefore, you would use those prefixes. And you don't write biaqua becaus...
- Thu Nov 14, 2013 12:22 pm
- Forum: Naming
- Topic: 16.27 aqua
- Replies: 1
- Views: 259
Re: 16.27 aqua
You're right! It's just rearranged differently to emphasize the part thats giving the electrons, which is oxygen.
- Wed Nov 13, 2013 11:48 pm
- Forum: Naming
- Topic: Cobalt vs. Cobaltate (#27)
- Replies: 2
- Views: 709
Re: Cobalt vs. Cobaltate (#27)
Add the ending -ate to to the cation if the overall charge is negative!
- Wed Nov 13, 2013 2:32 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: #71c in Chapter 10 HW Problems
- Replies: 2
- Views: 512
Re: #71c in Chapter 10 HW Problems
I think they're equivalent because if you increase the concentration, then the partial pressure increases as well.
Formula for pressure is
P= (n/v) RT
(n/v)= concentration
Formula for pressure is
P= (n/v) RT
(n/v)= concentration