Search found 144 matches

by Robert
Sat Mar 15, 2014 9:54 pm
Forum: *Organic Reaction Mechanisms in General
Topic: 2012 final Q5
Replies: 1
Views: 487

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.
by Robert
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: 352

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 ...
by Robert
Thu Mar 13, 2014 4:33 pm
Forum: *Cycloalkenes
Topic: Final 2013 6a Numbering a hexene
Replies: 8
Views: 2215

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.
by Robert
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: 589

Re: 2012 final Q3 (a)

That's how I determined the anode. Whichever reduction potentiometer was more negative/least positive!
by Robert
Thu Mar 13, 2014 3:58 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: Graphing
Replies: 1
Views: 580

Re: Graphing

The units should be in Kelvin because the formula k=e^-Ea/RT

The constant R is in J/mol*K
by Robert
Thu Mar 13, 2014 3:26 pm
Forum: General Rate Laws
Topic: Reaction Rates
Replies: 2
Views: 484

Re: Reaction Rates

The reaction rate is decreased by 1/4 because Cl2 is second order. (.5x)^2 = .25(x^2)
by Robert
Thu Mar 13, 2014 1:35 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: 7.19 Book Question
Replies: 2
Views: 448

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.
by Robert
Thu Mar 13, 2014 11:24 am
Forum: *Alkanes
Topic: Numbering based on alphabetical or substituent size?
Replies: 5
Views: 797

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
by Robert
Thu Mar 13, 2014 2:07 am
Forum: *Alkenes
Topic: Relative Stabilities of Alkenes/Alkynes
Replies: 3
Views: 1666

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...
by Robert
Sun Mar 09, 2014 5:52 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Residual entropy
Replies: 2
Views: 643

Re: Residual entropy

NO also has resonances; double and triple bonds
by Robert
Sun Mar 09, 2014 5:49 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Heat capacity
Replies: 2
Views: 543

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
by Robert
Sat Mar 08, 2014 3:01 pm
Forum: *Constitutional and Geometric Isomers (cis, Z and trans, E)
Topic: Final 2011 Q7B
Replies: 3
Views: 641

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...
by Robert
Wed Mar 05, 2014 10:29 am
Forum: Administrative Questions and Class Announcements
Topic: *Additional Forum Features*
Replies: 25
Views: 4028

Re: *Additional Forum Features*

It'd be cool if we could "like" or mark a post as "helpful."
by Robert
Tue Mar 04, 2014 4:07 pm
Forum: *Alkenes
Topic: Naming alkene along with halogen
Replies: 2
Views: 506

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.
by Robert
Tue Mar 04, 2014 3:34 pm
Forum: *Alkanes
Topic: naming
Replies: 1
Views: 333

Re: naming

Edit: in the hw problems and the practice finals, it seems both are accepted unless specified.
by Robert
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: 4457

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.
by Robert
Mon Mar 03, 2014 10:48 pm
Forum: *Chem3D
Topic: Why do bond angle values change?
Replies: 4
Views: 2056

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...
by Robert
Sat Mar 01, 2014 8:06 pm
Forum: *Alkanes
Topic: 33
Replies: 1
Views: 359

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.
by Robert
Thu Feb 27, 2014 11:27 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: to determine the overall reaction
Replies: 6
Views: 1614

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.
by Robert
Tue Feb 25, 2014 11:00 pm
Forum: *Enzyme Kinetics
Topic: Rate constant
Replies: 2
Views: 582

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.
by Robert
Tue Feb 25, 2014 10:47 pm
Forum: First Order Reactions
Topic: Quiz 2 2014 Prep: Question 4
Replies: 2
Views: 731

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...
by Robert
Tue Feb 25, 2014 10:43 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: rate constant
Replies: 1
Views: 528

Re: rate constant

k= Ae^-Ea/RT

Here the rate constant changes as a function of temperature. Ea is also a constant*
by Robert
Tue Feb 25, 2014 11:01 am
Forum: General Rate Laws
Topic: Rate law with intermediates
Replies: 3
Views: 815

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...
by Robert
Sat Feb 22, 2014 7:39 pm
Forum: First Order Reactions
Topic: Question 14.23 Integrated Rate Law
Replies: 2
Views: 499

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)
by Robert
Sat Feb 22, 2014 2:08 pm
Forum: General Rate Laws
Topic: Rate law with intermediates
Replies: 3
Views: 815

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...
by Robert
Sat Feb 22, 2014 1:12 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: to determine the overall reaction
Replies: 6
Views: 1614

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.
by Robert
Sat Feb 22, 2014 12:15 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: to determine the overall reaction
Replies: 6
Views: 1614

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...
by Robert
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: 445

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)
by Robert
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: 318

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...
by Robert
Mon Feb 10, 2014 6:23 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 2008 5A
Replies: 1
Views: 235

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...
by Robert
Sat Feb 08, 2014 5:32 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Winter 2009 midterm Question 1
Replies: 1
Views: 361

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 ...
by Robert
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: 1062

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 ...
by Robert
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: 481

Re: 13.103

That was very helpful! I appreciate your comprehensive explanation.
by Robert
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: 481

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 ...
by Robert
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: 395

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...
by Robert
Fri Jan 31, 2014 4:33 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 13.11
Replies: 4
Views: 605

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?
by Robert
Fri Jan 31, 2014 3:30 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 13.11
Replies: 4
Views: 605

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.
by Robert
Mon Jan 27, 2014 7:49 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 10.109
Replies: 1
Views: 1322

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...
by Robert
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: 368

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.
by Robert
Sat Jan 25, 2014 12:14 pm
Forum: Phase Changes & Related Calculations
Topic: Phase Change/heat curve Graphs
Replies: 2
Views: 535

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...
by Robert
Sat Jan 25, 2014 11:53 am
Forum: Administrative Questions and Class Announcements
Topic: Quiz 1
Replies: 3
Views: 626

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
by Robert
Fri Jan 24, 2014 9:18 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: HMW: 8.59 (a) and (c)
Replies: 3
Views: 632

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.

:)
by Robert
Fri Jan 24, 2014 6:19 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 8.103
Replies: 4
Views: 707

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...
by Robert
Thu Jan 23, 2014 12:30 pm
Forum: Phase Changes & Related Calculations
Topic: units
Replies: 2
Views: 386

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
by Robert
Thu Jan 23, 2014 12:28 pm
Forum: Phase Changes & Related Calculations
Topic: trouton's rule
Replies: 2
Views: 557

Re: trouton's rule

I know that organic means containing carbon, C. So any liquid hydrocarbons, alcohols, etc. look at appendix 2A, A16
by Robert
Thu Jan 23, 2014 12:26 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: entropy and volume
Replies: 7
Views: 922

Re: entropy and volume

Precisely!

dS = nR ln (P1/P2) --the formula when pressure is involved. notice that P initial is on top..
by Robert
Wed Jan 22, 2014 10:04 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 8.103
Replies: 4
Views: 707

Re: 8.103

Thank you!
by Robert
Wed Jan 22, 2014 7:00 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 8.103
Replies: 4
Views: 707

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...
by Robert
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: 475

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
by Robert
Tue Jan 21, 2014 11:32 am
Forum: Calculating Work of Expansion
Topic: Work of an isothermal reversible expansion
Replies: 4
Views: 627

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.
by Robert
Tue Jan 21, 2014 11:30 am
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: 8.43
Replies: 2
Views: 831

Re: 8.43

Thanks Iris!
by Robert
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: 799

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.
by Robert
Mon Jan 20, 2014 6:53 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: entropy and volume
Replies: 7
Views: 922

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.
by Robert
Thu Jan 16, 2014 11:06 am
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: 8.43
Replies: 2
Views: 831

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...
by Robert
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: 301

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...
by Robert
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: 2419

Re: Hmw: 7.53

Sorry I explained the whole problem; I interpreted your question incorrectly :(
by Robert
Sun Jan 12, 2014 12:25 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: 7.29 C
Replies: 3
Views: 462

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.
by Robert
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: 2419

Re: Hmw: 7.53

Treat them as separate ions. So Zn2+ and 2 Cl-
by Robert
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: 1453

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
by Robert
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: 2297

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...
by Robert
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: 628

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
by Robert
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: 2419

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...
by Robert
Wed Jan 08, 2014 6:44 pm
Forum: Phase Changes & Related Calculations
Topic: 7.17b
Replies: 2
Views: 518

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.
by Robert
Sat Dec 07, 2013 10:12 pm
Forum: *Indicators
Topic: What do you base the indicator on?
Replies: 1
Views: 720

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
by Robert
Sat Dec 07, 2013 9:44 pm
Forum: Empirical & Molecular Formulas
Topic: molecular formulas
Replies: 1
Views: 368

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.
by Robert
Sat Dec 07, 2013 6:55 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Unpaired Electrons
Replies: 1
Views: 4458

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...
by Robert
Sat Dec 07, 2013 6:41 pm
Forum: Molarity, Solutions, Dilutions
Topic: G. 25
Replies: 1
Views: 1093

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...
by Robert
Thu Dec 05, 2013 9:50 pm
Forum: *Titrations & Titration Calculations
Topic: 12.33
Replies: 2
Views: 421

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...
by Robert
Thu Dec 05, 2013 8:48 pm
Forum: *Titrations & Titration Calculations
Topic: Different equations
Replies: 1
Views: 278

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...
by Robert
Wed Dec 04, 2013 6:08 pm
Forum: Identifying Acidic & Basic Salts
Topic: 11.77
Replies: 1
Views: 371

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...
by Robert
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: 369

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...
by Robert
Wed Dec 04, 2013 5:58 pm
Forum: Bronsted Acids & Bases
Topic: How strong of an acid is HPO4 -?
Replies: 3
Views: 970

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
by Robert
Tue Dec 03, 2013 3:57 pm
Forum: Bronsted Acids & Bases
Topic: 11.19
Replies: 3
Views: 503

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.
by Robert
Mon Dec 02, 2013 10:08 pm
Forum: *Titrations & Titration Calculations
Topic: Reversal on Titration Curve
Replies: 2
Views: 446

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.
by Robert
Mon Dec 02, 2013 10:06 pm
Forum: Bronsted Acids & Bases
Topic: NH3+ + HCl ---> NH4Cl
Replies: 4
Views: 5375

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
by Robert
Mon Dec 02, 2013 8:55 pm
Forum: Polyprotic Acids & Bases
Topic: Determining what is a polyprotic acid/base
Replies: 5
Views: 789

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.
by Robert
Mon Dec 02, 2013 8:49 pm
Forum: *Making Buffers & Calculating Buffer pH (Henderson-Hasselbalch Equation)
Topic: 11.21a
Replies: 4
Views: 467

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.
by Robert
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: 23620

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...
by Robert
Mon Dec 02, 2013 12:05 pm
Forum: Lewis Acids & Bases
Topic: HCl
Replies: 4
Views: 737

Re: HCl

The species HCl is a Lewis acid because it acts as a proton donor, the proton being h+
by Robert
Sat Nov 30, 2013 6:28 pm
Forum: Conjugate Acids & Bases
Topic: 2008 final exam 6.c
Replies: 2
Views: 395

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.
by Robert
Wed Nov 27, 2013 11:54 pm
Forum: Bronsted Acids & Bases
Topic: 11.71
Replies: 2
Views: 345

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...
by Robert
Wed Nov 27, 2013 11:49 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: 11.75c
Replies: 4
Views: 489

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...
by Robert
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: 470

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.
by Robert
Tue Nov 26, 2013 11:28 am
Forum: Identifying Acidic & Basic Salts
Topic: 11.19
Replies: 8
Views: 1096

Re: 11.19

Metalloids are amphoteric
by Robert
Thu Nov 21, 2013 12:55 pm
Forum: Identifying Acidic & Basic Salts
Topic: 11.19
Replies: 8
Views: 1096

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
by Robert
Wed Nov 20, 2013 9:58 pm
Forum: Naming
Topic: Copper vs. Cupperate
Replies: 2
Views: 840

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.
by Robert
Tue Nov 19, 2013 7:28 pm
Forum: Naming
Topic: Am I alphabetizing Wrong
Replies: 5
Views: 836

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.
by Robert
Tue Nov 19, 2013 6:59 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Equilibrium calculations in the ICE Table
Replies: 3
Views: 519

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.,
by Robert
Thu Nov 14, 2013 8:50 pm
Forum: Naming
Topic: Order of Coordination Compound inside the brackets.
Replies: 2
Views: 463

Re: Order of Coordination Compound inside the brackets.

For your first question, you alphabetize. I2, nh3, then o2. Same rule applies to multiple anions.
by Robert
Thu Nov 14, 2013 8:42 pm
Forum: Shape, Structure, Coordination Number, Ligands
Topic: 16.31
Replies: 1
Views: 427

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...
by Robert
Thu Nov 14, 2013 8:35 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Cubic equations for x
Replies: 2
Views: 430

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)
by Robert
Thu Nov 14, 2013 8:31 pm
Forum: Naming
Topic: Cyanido vs. Isocyanido
Replies: 3
Views: 645

Re: Cyanido vs. Isocyanido

The one that is linked to the metal ion will be underlined.
by Robert
Thu Nov 14, 2013 8:30 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Homework 10.27
Replies: 2
Views: 434

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 :)
by Robert
Thu Nov 14, 2013 8:22 pm
Forum: Naming
Topic: Problem from textbook ch.16, 16.29 d
Replies: 3
Views: 599

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...
by Robert
Thu Nov 14, 2013 12:22 pm
Forum: Naming
Topic: 16.27 aqua
Replies: 1
Views: 249

Re: 16.27 aqua

You're right! It's just rearranged differently to emphasize the part thats giving the electrons, which is oxygen.
by Robert
Wed Nov 13, 2013 11:48 pm
Forum: Naming
Topic: Cobalt vs. Cobaltate (#27)
Replies: 2
Views: 644

Re: Cobalt vs. Cobaltate (#27)

Add the ending -ate to to the cation if the overall charge is negative!
by Robert
Wed Nov 13, 2013 2:32 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: #71c in Chapter 10 HW Problems
Replies: 2
Views: 437

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

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