Search found 89 matches
- Thu Mar 14, 2019 1:40 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Test 2, Question 6 part a
- Replies: 4
- Views: 486
Re: Test 2, Question 6 part a
But because it says it goes from neutral to second oxidation, does that not mean you have to switch the signs?
- Thu Mar 14, 2019 11:15 am
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: 6th edition, 15.89
- Replies: 1
- Views: 225
6th edition, 15.89
The rate law of the reaction 2 NO(g) + 2 H2(g) -> N2(g) + 2 H2O(g) is Rate = k[NO]2[H2], and the mechanism that has been proposed is Step 1 NO + NO -> N2O2 Step 2 N2O2 + H2 -> N2O + H2O Step 3 N2O + H2 -> N2 + H2O (a) Which step in the mechanism is likely to be rate determining? Explain your answer....
- Thu Mar 14, 2019 10:33 am
- Forum: Experimental Details
- Topic: 6th edition, 15.87
- Replies: 1
- Views: 745
6th edition, 15.87
The hydrolysis of sucrose (C12H22O11) produces fructose and glucose: C12H22O11(aq) + H2O(l) -> C6H12O6(glucose, aq) + C6H12O6(fructose, aq). Two mechanisms are proposed for this reaction: (i) Step 1 C12H22O11 -> C6H12O6 + C6H10O5 (slow) Step 2 C6H10O5 + H2O -> C6H12O6 (fast) (ii) C12H22O11 + H2O -> ...
- Wed Mar 13, 2019 9:14 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Test 2, Question 6 part a
- Replies: 4
- Views: 486
Test 2, Question 6 part a
Rank the following species in order of increasing reducing power going from their neutral to second oxidation state: Pb, Cd, Mn
What is the answer for this question? I got: Mn < Cd < Pb but I got it incorrect so what is the correct answer?
What is the answer for this question? I got: Mn < Cd < Pb but I got it incorrect so what is the correct answer?
- Wed Mar 13, 2019 2:21 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Organizing Reducing/Oxidizing Power
- Replies: 2
- Views: 422
Re: Organizing Reducing/Oxidizing Power
Does anyone know the specific answer to the question regarding Pb, Cd, Mn? I got Mn < Cd < Pb, but I was marked wrong. Why is that?
- Wed Mar 13, 2019 2:19 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: T, Ea, and A's effect on K
- Replies: 1
- Views: 289
T, Ea, and A's effect on K
Can someone explain the effect of T, Ea, and A's effect on K from both a conceptual approach and a mathematical approach using the equation? Also, in the textbook, they state: "the higher the activation energy, the stronger is the temperature dependence of the rate constant". Can someone e...
- Tue Mar 12, 2019 11:29 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Organizing Reducing/Oxidizing Power
- Replies: 2
- Views: 422
Organizing Reducing/Oxidizing Power
I know the pattern to organize species in reducing/oxidizing power, but when you do, does the wording or specification matter? For example, on test 2, one of the questions was: "rank the following species (Pb, Cd, Mn) in order of increasing reducing power (so more negative, the stronger) going ...
- Mon Mar 11, 2019 4:45 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: mmol vs mol
- Replies: 6
- Views: 1071
mmol vs mol
Does it matter whether our answer is given in mmol or mol? Because for example, in the 6th edition, problem 17, the rate constant and rate answers are given in mmol whereas in problem 19, the rate constant and rate answers are given in mol. Will we get marked off if we give either on the final?
- Sat Mar 09, 2019 10:16 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: 6th edition, 14.55
- Replies: 2
- Views: 345
Re: 6th edition, 14.55
What do you mean by it has lower energy requirements to oxidize?
- Sat Mar 09, 2019 9:52 pm
- Forum: Interesting Applications: Rechargeable Batteries (Cell Phones, Notebooks, Cars), Fuel Cells (Space Shuttle), Photovoltaic Cells (Solar Panels), Electrolysis, Rust
- Topic: 6th edition, 14.119
- Replies: 1
- Views: 485
6th edition, 14.119
One stage in the extraction of gold from rocks involves dissolving the metal from the rock with a basic solution of sodium cyanide that has been thoroughly aerated. This stage results in the formation of soluble Au(CN)2- ions. The next stage is to reduce gold to the metal by the addition of zinc dus...
- Sat Mar 09, 2019 6:00 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: 6th edition, 14.55
- Replies: 2
- Views: 345
6th edition, 14.55
A 1.0 m NiSO4(aq) solution was electrolyzed by using inert electrodes. Write (a) the cathode reaction; (b) the anode reaction. (c) With no overpotential or passivity at the electrodes, what is the minimum potential that must be supplied to the cell for the onset of electrolysis? I'm not sure how the...
- Thu Mar 07, 2019 3:30 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: types of units for K in nernst equation
- Replies: 1
- Views: 297
types of units for K in nernst equation
Do the units all need to be the same when calculating K for Nernst equation. For example, if one of the reactant was in mol.L-1 and another was atm or bar, do we need to convert one to the other? (See 14.37 as an example)
- Thu Mar 07, 2019 3:15 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: 6th edition, 14.37
- Replies: 1
- Views: 221
6th edition, 14.37
Determine the potential of each of the following cells: (a) Pt(s) | H2(g, 1.0 bar) | HCl(aq, 0.075 m) || HCl(aq, 1.0 mol.L.-1) | H2(g, 1.0 bar) | Pt(s) In the solutions manual, they got 2H+ + 2e- -> H2 (g) as the equation for the cathode and H2 (g) -> 2H+ (aq) + 2e- as the anode. How did that happen...
- Thu Mar 07, 2019 12:12 am
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: 6th edition, 14.35
- Replies: 1
- Views: 277
6th edition, 14.35
Determine the equilibrium constants for the following reactions: (a) Mn(s) + Ti2+ (aq) <-> Mn2+(aq) + Ti(s) In the solutions manual, how did they obtain the equation: at 25C lnK = nE/0.02568 V. Also, is there a way to just plug in numbers into the original formula, lnK = nFE/RT? I'm having trouble b...
- Thu Feb 28, 2019 4:39 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: 6th edition, 14.33
- Replies: 1
- Views: 239
6th edition, 14.33
a) The standard Gibbs free energy of formation of Tl3+(aq) is +215 kJ.mol-1 at 25C. Calculate the standard potential of the Tl3+/Tl couple. (b) Will Tl+ disproportionate in aqueous solution Can someone explain how a redox couple works and for b) I found delta G in the end, but how does delta G for T...
- Thu Feb 28, 2019 7:08 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 6th edition, 14.17
- Replies: 2
- Views: 301
6th edition, 14.17
(a) Write balanced half-reactions for the redox reaction of an acidified solution of potassium permanganate and iron(II) chloride.
How would we start off this question? I'm not sure where they got Fe2+ -> Fe3+ from and MnO4- -> Mn2+
How would we start off this question? I'm not sure where they got Fe2+ -> Fe3+ from and MnO4- -> Mn2+
- Thu Feb 28, 2019 6:25 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 6th edition, 14.15b
- Replies: 1
- Views: 203
6th edition, 14.15b
Write the half-reactions and devise a galvanic cell (write a cell diagram) to study each of the following reactions: (b) H+ (aq) + OH-(aq) -> H2O(l), the Brønsted neutralization reaction Why can we not use the equations: 2 H+ + 2 e- -> H2 , H2 + 2 OH- -> 2 H2O + 2 e- to get the redox reaction? Also,...
- Wed Feb 27, 2019 9:21 pm
- Forum: Balancing Redox Reactions
- Topic: 6th edition, 14.5d
- Replies: 1
- Views: 350
6th edition, 14.5d
Balance each of the following skeletal equations by using oxidation and reduction half-reactions. All the reactions take place in basic solution. Identify the oxidizing agent and reducing agent in each reaction. (d) Reaction of elemental phosphorus to form phosphine, PH3, a poisonous gas with the od...
- Wed Feb 27, 2019 8:14 pm
- Forum: Balancing Redox Reactions
- Topic: 6th edition, 14.3b
- Replies: 1
- Views: 1345
6th edition, 14.3b
Balance each of the following skeletal equations by using oxidation and reduction half-reactions. All the reactions take place in acidic solution. Identify the oxidizing agent and reducing agent in each reaction. (a) Reaction of thiosulfate ion with chlorine gas: Cl2 (g) + S2O3 2- (aq) -> Cl- (aq) +...
- Wed Feb 27, 2019 7:32 pm
- Forum: Balancing Redox Reactions
- Topic: 6th Edition, 14.1 [ENDORSED]
- Replies: 1
- Views: 291
6th Edition, 14.1 [ENDORSED]
The following redox reaction is used in acidic solution in the Breathalyzer test to determine the level of alcohol in the blood: H+ (aq) + Cr2O7 2- (aq) + C2H5OH (aq) -> Cr3+ (aq) + C2H4O (aq) + H2O (l) (a) Identify the elements undergoing changes in oxidation state and indicate the initial and fina...
- Fri Feb 22, 2019 10:44 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: delta S= q(rev)/ T
- Replies: 5
- Views: 2222
Re: delta S= q(rev)/ T
Note that the equation contains q(rev), where the subscript "rev" on q signified that the energy must be transferred reversibly and in a reversible reaction, temperature is constant. You can also use this equation to work out how to calculate the entropy of vaporization or the entropy of f...
- Thu Feb 21, 2019 11:58 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Negative ∆G means spontaneous reaction?
- Replies: 5
- Views: 7079
Negative ∆G means spontaneous reaction?
Why does a negative ∆G correlate with a spontaneous reaction? I thought that ∆G was the change in gibbs free energy so why is the reaction spontaneous when the change is negative?
- Thu Feb 21, 2019 11:06 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Gibbs Free Energy vs Entropy
- Replies: 4
- Views: 800
Gibbs Free Energy vs Entropy
What is the difference between Gibbs Free Energy and entropy and why is gibbs free energy so important??
- Tue Feb 12, 2019 11:24 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Whether PV is significant
- Replies: 2
- Views: 294
Re: Whether PV is significant
Does the entire reaction have to contain all gases to be significant and vice versa? So if the reaction have gas as reactants and solid as products, is PdeltaV significant or nonsignificant?
- Tue Feb 12, 2019 7:45 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Whether PV is significant
- Replies: 2
- Views: 294
Whether PV is significant
For the first law of thermodynamics in our notes, there is this part about constant volume and constant pressure. For constant pressure, can someone explain the significance of "for a reaction at constant pressure that involves changes in the number of moles of gas, then PdeltaV is significant ...
- Tue Feb 12, 2019 7:18 pm
- Forum: Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy
- Topic: 6th edition 9.75
- Replies: 2
- Views: 576
Re: 6th edition 9.75
Use exponential rules and bring Avogadro's number down in front of ln. Also, is this the wrong question? I don't think you need to use Avogadro's number in 9.75.
- Tue Feb 12, 2019 4:13 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 6th edition 9.25
- Replies: 1
- Views: 232
Re: 6th edition 9.25
You have to set it to the power of avogadro's number: 6^(6.023 x 10^23). If you put this into the calculator, it will give you an error, but you can manipulate it with exponent rules and move it down in front of ln.
- Tue Feb 12, 2019 1:32 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: temp units
- Replies: 3
- Views: 367
Re: temp units
I believe that the units for entropy/entropy change are always in J.K-1 so the units should be in Kelvin. I haven't seen entropy in Celsius yet.
- Sat Feb 09, 2019 12:13 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 6th Edition, 8.99
- Replies: 1
- Views: 233
6th Edition, 8.99
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 temperatu...
- Thu Feb 07, 2019 10:41 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 7th Edition 4.15
- Replies: 2
- Views: 498
Re: 7th Edition 4.15
Can you explain why you want to find the limiting reactant? What's different about this problem from others where you don't have to find the limiting reactant?
- Thu Feb 07, 2019 9:09 pm
- Forum: Phase Changes & Related Calculations
- Topic: 6th Edition, 8.91
- Replies: 1
- Views: 270
6th Edition, 8.91
In 1750 Joseph Black performed an experiment that eventually led to the discovery of enthalpies of fusion. He placed two 150.-g samples of water at 0.00 C (one ice and one liquid) in a room kept at a constant temperature of 5.00 C. He then observed how long it took for each sample to warm to its fin...
- Thu Feb 07, 2019 1:21 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 6th edition, 8.53
- Replies: 2
- Views: 219
Re: 6th edition, 8.53
But in the answer key, they only used 1.40 g. It never mentions 1.00 mol CO. So why do we disregard 1.00 mol?
- Thu Feb 07, 2019 10:34 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 6th edition, 8.53
- Replies: 2
- Views: 219
6th edition, 8.53
8;53 The reaction of 1.40 g of carbon monoxide with excess water vapor to produce carbon dioxide and hydrogen gases in a bomb calorimeter causes the temperature of the calorimeter assembly to rise from 22.113 C to 22.799 C. The calorimeter assembly is known to have a total heat capacity of 3.00 kJ·(...
- Tue Feb 05, 2019 9:42 pm
- Forum: Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy
- Topic: Determining the Number of Orientations/Degeneracy & Microstates
- Replies: 3
- Views: 658
Re: Determining the Number of Orientations/Degeneracy & Microstates
In the example he used in class, what does he mean by "2 particles both in one of two states, W = 4 (or 2^2)" and "1 particle in one of two states, W = 2 (or 2^1)". Was there 4 microstates or 2?
- Tue Feb 05, 2019 9:39 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Monday: Derivation of Isothermal, Reversible Expansion
- Replies: 2
- Views: 428
Monday: Derivation of Isothermal, Reversible Expansion
In Monday's Lecture Notes, after talking about entropy, the slide moved onto deriving the equation for isothermal, reversible expansion. There were a bunch of equations after "need to know". Can someone explain the importance of those equations, especially ∆U = 3/2nR∆T = 0. Also, what does...
- Tue Feb 05, 2019 9:21 pm
- Forum: Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy
- Topic: Determining the Number of Orientations/Degeneracy & Microstates
- Replies: 3
- Views: 658
Determining the Number of Orientations/Degeneracy & Microstates
Is degeneracy and microstates the same thing since degeneracy is W and microstates is also represented by W. Also, I know that W = (number of possible positions/orientations)^(number of particles), but how do you determine the base, which is the number of possible positions/orientations? Can someone...
- Mon Feb 04, 2019 5:08 pm
- Forum: Calculating Work of Expansion
- Topic: Units for Work
- Replies: 2
- Views: 308
Units for Work
Is Pa.m^3 the only expanded "unit" that is equivalent to Joules? Are there other units we should know about??
- Tue Jan 29, 2019 4:54 pm
- Forum: Phase Changes & Related Calculations
- Topic: Units: Joules vs kJ
- Replies: 5
- Views: 562
Units: Joules vs kJ
What is the common preference for the units for ΔHvap, ΔHfus, ΔHsub? Is it joules or kilojoules? Because in problem 8.37, the answer is in kilojoules, but would we be incorrect if we answered in joules?
- Tue Jan 29, 2019 4:40 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Method 2: Using bond enthalpies
- Replies: 3
- Views: 380
Method 2: Using bond enthalpies
Can someone explain how the method of using bond enthalpies work in a calculation?
- Tue Jan 29, 2019 4:25 pm
- Forum: Calculating Work of Expansion
- Topic: Work and Internal Energy
- Replies: 1
- Views: 196
Work and Internal Energy
Are we supposed to know the relationship between work and internal energy already? It doesn't seem like we have gone over work, internal energy, or the first law of thermodynamics much in lecture. For example, on 8.3 in the 6th edition, it asks to calculate the work done in the compression and inter...
- Tue Jan 29, 2019 4:21 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 6th Edition, 8.57
- Replies: 1
- Views: 282
6th Edition, 8.57
Determine the reaction enthalpy for the hydrogenation of ethyne to ethane, C2H2(g) + 2 H2(g) -> C2H6(g), from the following data: ΔHc(C2H2, g) = -1300. kJ·mol-1, ΔHc(C2H6, g) = -1560. kJ·mol-1, ΔHc(H2, g) = -286 kJ·mol-1. Can someone explain how to do this?
- Fri Jan 25, 2019 10:07 am
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: Homework 6th edition 12.55.D
- Replies: 1
- Views: 445
Re: Homework 6th edition 12.55.D
The difference in pH can be explained by the type of bond and strength of that bond between the oxygen and hydrogen. Since double bonds are stronger than single bonds, these acids would be less likely to give up a hydrogen, which means there's a smaller concentration of hydronium ions, thus, a great...
- Tue Jan 22, 2019 6:02 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Outline 1, last bullet
- Replies: 2
- Views: 335
Re: Outline 1, last bullet
Le Chatelier's Principle states: chemical reactions adjust so as to minimize the effect of any changes. Adding or Removing Reagents: [CONCENTRATION] -Adding reactants: there is more reactants than products so in order to reach equilibrium again, the reaction will proceed towards the right and more p...
- Mon Jan 21, 2019 12:26 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Applied Exercises 12.131
- Replies: 1
- Views: 295
Re: Applied Exercises 12.131
a) Muscles work hard -> no oxygen -> produces lactic acid. If you think about what happens when the concentration of oxygen decreases, based on what we learned on the response to change of concentration, more oxygen will be produced to replace the oxygen that has been decreased. If more oxygen is pr...
- Mon Jan 21, 2019 12:20 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Error in Solutions manual [ENDORSED]
- Replies: 1
- Views: 1019
Re: Error in Solutions manual [ENDORSED]
There is no error in the solutions manual. If you have calculated correctly, the Ka values for the acids should be: (7.6 x 10^-3) for H3PO4, (1.0 x 10^-2) for H3PO3, (3.5 x 10^-3) for H2SeO3, and (1.2 x 10^-2) for HSeO4-. As you have stated, the larger the Ka value, the stronger the acid. This also ...
- Wed Jan 16, 2019 4:59 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Weak Acids and Bases
- Replies: 6
- Views: 580
Re: Weak Acids and Bases
What if the acid was a strong acid (x is greater than 10^-7) instead of a weak acid? Would you still add it to 10^-7?
- Wed Jan 16, 2019 4:13 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: relationship between Ka, Kb, and its ability to donate
- Replies: 3
- Views: 280
relationship between Ka, Kb, and its ability to donate
Is this correct: The smaller the value of Ka, the weaker the ability of the acid to donate a proton, but the greater the value of pKa. This also means the acid is very weak. Similarly, the smaller the value of Kb, the weaker the ability of the base to accept a proton, but the greater value of pKb. T...
- Tue Jan 15, 2019 11:17 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 6th edition 12.25
- Replies: 1
- Views: 235
Re: 6th edition 12.25
To calculate the initial molarity of Ba(OH)2, [remember molarity= mol/volume (L)], convert 0.43g into mol and divide by 0.100 because that's the volume: 0.43g (1mol/171.344g)(1/100) = 0.025 M Ba(OH)2 Then, if you know that it's in an aqueous solution, Ba(OH)2 will just dissociate and become ions: Ba...
- Mon Jan 14, 2019 3:29 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 6th Edition 11.89
- Replies: 3
- Views: 251
Re: 6th Edition 11.89
I did do that, but the answer in the answer key is: (5/100)(10/100)^2/(18/100)^2. Can someone explain why this is?
- Mon Jan 14, 2019 2:51 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 6th Edition 11.89
- Replies: 3
- Views: 251
6th Edition 11.89
The following plot shows how the partial pressures of reactant and products vary with time for the decomposition of compound A into compounds B and C. All three compounds are gases. Use this plot to do the following: (a) Write a balanced chemical equation for the reaction. (b) Calculate the equilibr...
- Mon Jan 14, 2019 2:37 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: HW Q 6th ed. 11.53
- Replies: 5
- Views: 1014
Re: HW Q 6th ed. 11.53
If you have already calculated the concentration of H2 at equilibrium, which should be 0.053, then you know that the change in concentration of H2 should be 0.08. Note that how I got 0.08 was 60% (.60) of the gas reacted so .60 x 0.400 which equals .24 and to find concentration, divide by 3, getting...
- Mon Jan 14, 2019 9:25 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 7th edition 5I.13
- Replies: 1
- Views: 146
Re: 7th edition 5I.13
So for part B, the equation should be: F2(g) -> 2(F) BUT, the equilibrium constant (Kc) is 1.2 x 10^-4 Essentially, your ICE table for part B should be very similar to your ICE table for part A. Looks something like this: F2 F I 0.001 0 C -x +x E 0.001-x x Kc= [F]^2 /[F2] = (2X)^2/(0.001-X) = 4X^2/0...
- Wed Jan 09, 2019 5:07 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Brackets vs P For Equilibrium Constants
- Replies: 4
- Views: 207
Re: Brackets vs P For Equilibrium Constants
Thank you! But as a follow up question, in the 6th edition, number 11.13 part a, it asks to write the reaction quotient for: 2 BCl3(g) + 2 Hg(l) -> B2Cl4(s) + Hg2Cl2(s) and in the answer key, the answer is given as: 1/ P BCl3 ^2. There was no "partial pressure" or "concentration"...
- Wed Jan 09, 2019 5:02 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Brackets vs P For Equilibrium Constants
- Replies: 4
- Views: 207
Brackets vs P For Equilibrium Constants
In an equilibrium constant expression (to find the equilibrium constant), it's [products]/ [reactants] (the concentration of products over the concentration of reactants), but if something is aqueous, do we use the brackets and if something is a gas, then we use partial pressure? For example, P4S10(...
- Wed Jan 09, 2019 4:55 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Q
- Replies: 6
- Views: 377
Re: Q
How did you get 1 in the numerator? I know that the molar concentration of a pure substance (solid or liquid) does not change in a reaction so they are not included, but where did you get the 1 from?
- Mon Jan 07, 2019 3:38 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Homework Problem 11.3 for 6th Edition
- Replies: 3
- Views: 275
Re: Homework Problem 11.3 for 6th Edition
So, does that mean if you wrote it like this:
Kc = [C2H4Cl2]^2 [H2O]^2 / [C2H4]^2 [O2] [HCl]^4
is this still correct because you wouldn't have to write P (partial pressure)?
Kc = [C2H4Cl2]^2 [H2O]^2 / [C2H4]^2 [O2] [HCl]^4
is this still correct because you wouldn't have to write P (partial pressure)?
- Sun Dec 09, 2018 10:05 am
- Forum: Bronsted Acids & Bases
- Topic: Why is this not a proton transfer. (7th ed. 6A.9)
- Replies: 1
- Views: 514
Re: Why is this not a proton transfer. (7th ed. 6A.9)
You are transferring the NH to CH3COOH to make CH3CONH2 and O to NH3 to make H2O together, not necessarily a H+ so it wouldn't count.
- Fri Dec 07, 2018 10:39 pm
- Forum: Polarisability of Anions, The Polarizing Power of Cations
- Topic: Least to Most Polarizable/Polarizing Power
- Replies: 3
- Views: 2067
Re: Least to Most Polarizable/Polarizing Power
Why is S2- < P3-? and why is Mg2+ < Li+?
- Fri Dec 07, 2018 2:07 pm
- Forum: Amphoteric Compounds
- Topic: Oxalate bidentate
- Replies: 4
- Views: 2511
Re: Oxalate bidentate
Oxalate is bidentate because there are 2 places that can bond to the metal, M. In oxalate, the formula is C2O4 and two of the oxygens will bind to the metal, which makes it bidentate, unlike monodentate, where only one atom will bond with the metal. In the pictures below, M- metal, and you can see t...
- Fri Dec 07, 2018 2:02 pm
- Forum: Polarisability of Anions, The Polarizing Power of Cations
- Topic: Least to Most Polarizable/Polarizing Power
- Replies: 3
- Views: 2067
Least to Most Polarizable/Polarizing Power
Order these anions in order from least to most polarizable: Br-, S2-, O2-, P3-
Order these cations in order from least to most polarizing power: Li+, Mg2+, Na+, Be2+
Order these cations in order from least to most polarizing power: Li+, Mg2+, Na+, Be2+
- Thu Dec 06, 2018 10:03 pm
- Forum: Properties & Structures of Inorganic & Organic Acids
- Topic: Fundamentals J.7b 6th Edition
- Replies: 1
- Views: 129
Fundamentals J.7b 6th Edition
For Question J.7. b, I got the answer:
HNO2(aq) + ZnOH (aq) -> H2O (l) +ZnNO2, but it does not match the answer in the solutions manual. Can someone explain why this is wrong?
HNO2(aq) + ZnOH (aq) -> H2O (l) +ZnNO2, but it does not match the answer in the solutions manual. Can someone explain why this is wrong?
- Fri Nov 30, 2018 8:11 pm
- Forum: Dipole Moments
- Topic: Induced Dipole - Induced Dipole
- Replies: 3
- Views: 477
Re: Induced Dipole - Induced Dipole
No, induced-induced dipole is between 2 nonpolar molecules whereas dipole-dipole is between two polar molecules where the partial negative would be attracted with the partial positive charge of the other molecule. In an induced-induced dipole, since the molecule is non-polar, that means one molecule...
Re: Naming
When you name the transition metals, you would use the Roman numerals after the name of the transition metal. The Roman numeral matches the charge of the ion, so for example, if your ion was Fe 2+, then you would write it with the Roman numeral two in parenthesis after the name, like Iron (II).
- Fri Nov 30, 2018 8:04 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Seesaw
- Replies: 13
- Views: 2095
Re: Seesaw
The angles for seesaw is <90 and <120 because for a normal trigonal bipyramidal, the bond angles are 90 and 120 so if you replaced one of the atoms with a lone pair, then it would repel the others, causing the bond angles to become smaller.
- Wed Nov 21, 2018 1:58 pm
- Forum: Hybridization
- Topic: Purpose of hybridization
- Replies: 4
- Views: 398
Purpose of hybridization
What is the purpose of hybridization? I know VSEPR allows us to figure out the actual shape of a molecule accurately, but what does hybridization do and why is it important?
- Wed Nov 21, 2018 1:41 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Shape of H2O
- Replies: 1
- Views: 408
Re: Shape of H2O
Remember that the bond angles in a tetrahedral is not 90 degrees. The lewis structure is different from the vsepr geometry so, putting the lone pairs opposite each other would not make it linear. You can see, at the picture attached, that there really isn't any "opposite". Also, according ...
- Wed Nov 21, 2018 1:31 pm
- Forum: Electronegativity
- Topic: octet expansion
- Replies: 2
- Views: 524
Re: octet expansion
The reason to the octet expansion involving d-orbitals is because if n=2, the second row of the periodic table, the electron configuration would only be s2p6, which involves no d-orbitals, therefore, aren't allowed for octet expansion. On the other hand, if n=3, you can recall that l would equal to ...
- Fri Nov 16, 2018 5:40 pm
- Forum: Dipole Moments
- Topic: Definition of a dipole
- Replies: 2
- Views: 177
Definition of a dipole
Can someone explain the definition of a dipole or an electric dipole moment and does this have anything to do with a molecule being polar/non-polar?
- Fri Nov 16, 2018 5:27 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Octahedral
- Replies: 4
- Views: 546
Re: Octahedral
For 6 electron pairs:
Octahedral: AX6
Square pyramidal: AX5E
Square planar: AX4E2
Octahedral: AX6
Square pyramidal: AX5E
Square planar: AX4E2
- Fri Nov 16, 2018 5:25 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Single Regions?
- Replies: 8
- Views: 800
Re: Single Regions?
What is meant by "single region" is that there is this one area where this bond is happening, whether it be a single, double, or triple bond. Like for example, if you have a double bond, it doesn't mean you have 2 regions were a bond is happening. It's only one region where two atoms are b...
- Fri Nov 09, 2018 1:34 pm
- Forum: Quantum Numbers and The H-Atom
- Topic: Vocabulary confusion
- Replies: 2
- Views: 577
Re: Vocabulary confusion
The principal quantum number is n, which can be 1, 2, 3... (all integers from 1) and they represents the size of the orbital and energy. Magnetic is denoted as ml, which is basically -l....0...+l and it ml represents the orbital orientation. Note: l is called angular and can be integers starting fro...
- Fri Nov 09, 2018 1:29 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: Test 2 1b
- Replies: 2
- Views: 518
Re: Test 2 1b
If there is uncertainty in momentum, then there is uncertainty in wavelength because wavelength equal h/(uncertainty of momentum [p]) results in uncertainty in wavelength = h/ (uncertainty in momentum). If the uncertainty in momentum increases, the uncertainty in its wavelength decreases. If uncerta...
- Fri Nov 09, 2018 1:22 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: VSEPR of H2O
- Replies: 1
- Views: 197
VSEPR of H2O
Since BeCl2 is linear and the central atom has 2 bonding pair, does that mean H2O is also linear and has 180 degree bond angle? If not, why?
- Thu Nov 01, 2018 5:31 pm
- Forum: Ionic & Covalent Bonds
- Topic: 6th Edition, 3.5
- Replies: 1
- Views: 372
Re: 6th Edition, 3.5
How I looked at these types of problems is I first wrote the electron configuration out. Then, I rearranged it according to the charge and number. For (b), Bi3+: if you look at the periodic table, the noble gas notation is [Xe]. From there, I just list it from left to right: so, first is 6s2. When y...
- Thu Nov 01, 2018 5:13 pm
- Forum: Lewis Structures
- Topic: 6th Edition, Problem 3.33
- Replies: 1
- Views: 162
6th Edition, Problem 3.33
3.33: Write the Lewis structure of (a)CCl4; (b)COCl2; (c) ONF; (d)NF3 For 3.33 c, Nitrogen is the central atom because it has the lowest electron affinity, but does it matter if Oxygen and Fluorine are placed on the left or right? For example, if Oxygen was on the left versus right, is one more corr...
- Thu Nov 01, 2018 4:15 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Ag+ Configuration
- Replies: 5
- Views: 1929
Re: Ag+ Configuration
I think Ag is like an exception, where you remove one electron from the s orbital and it makes the d orbital full, so d10. Cu (copper) and Cr (chromium) follow the same pattern. So, Ag+ is originally [Kr] 5s24d9, but you remove one from 5s2 to make the 4d orbital full, so it becomes 5s1 4d10 and the...
- Sun Oct 28, 2018 10:15 pm
- Forum: Trends in The Periodic Table
- Topic: Lewis Structures
- Replies: 7
- Views: 727
Re: Lewis Structures
I believe it doesn't matter if the dots are left or right, but it DOES matter if you have 2 paired electron. If there is 2 paired electron, you need to make sure that they are shown in your diagram/drawing of your lewis dot structure. For example, Mg has 2 paired electrons. So you wouldn't put one a...
- Sun Oct 28, 2018 10:07 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Wave Function Squared
- Replies: 2
- Views: 288
Re: Wave Function Squared
Since wave function squared means the probability of finding an electron, does that mean the wave function of a nodal plane is 0? Since you won't find an electron in a nodal plane?
- Mon Oct 22, 2018 5:00 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: question 42 on post assessment
- Replies: 2
- Views: 425
Re: question 42 on post assessment
In the atomic spectra module (video portion), at one point (at 26.04), Professor Lavelle introduces a formula that he derives, which involves the frequency and 2 energy levels. The formula is: V = R [(1/n1^2) - (1/n2^2)]. So, you know that the frequency (v) = 1.14 x 10^14 Hz. The energy level that i...
- Thu Oct 18, 2018 11:42 pm
- Forum: Photoelectric Effect
- Topic: Photoelectric Effect Post-Assessment Module #18
- Replies: 2
- Views: 155
Re: Photoelectric Effect Post-Assessment Module #18
The answer is E) both C and D because the work function (also called your threshold energy) indicates how much energy is needed to eject an electron. (Think of it like a threshold, you need to reach a certain level for something to occur). Therefore, if you have more than enough, that is E(photon) >...
- Thu Oct 18, 2018 11:34 pm
- Forum: Einstein Equation
- Topic: Values of Each Symbol
- Replies: 3
- Views: 452
Re: Values of Each Symbol
I believe professor Lavelle said that there will be a cover page full of constants, like the speed of light, etc. on the tests so they will be given to you. It should be the very front page and the constants should be listed, but they won't tell you WHICH constant to use for which problem so you sho...
- Thu Oct 18, 2018 4:22 pm
- Forum: Properties of Light
- Topic: Atomic Spectrum and Series
- Replies: 4
- Views: 177
Re: Atomic Spectrum and Series
Wait, can someone elaborate what it means by a close range of wavelengths? These series are different because for the Balmer series, the energized electrons fall down to n=2, whereas the lyman series falls down to n=1 right? Also, in the answer key, it says "in each of these series, the princip...
- Wed Oct 17, 2018 4:43 pm
- Forum: Photoelectric Effect
- Topic: Photoelectric Effect Post-Module Assessment
- Replies: 2
- Views: 266
Re: Photoelectric Effect Post-Module Assessment
You know the energy which is 7.807 x 10^-19 J. If you want to find the wavelength, use the formula λ=(hc)/E to find the wavelength: λ=(6.626 x 10^-34 J.s^-1)(3.00 x 10^8 m.s^-1) / (7.807 x 10^-19 J) = 2.54 x 10^-7 m or 254 nm OR you can take the longer step, which is first finding out the frequency,...
- Fri Oct 12, 2018 12:21 am
- Forum: Molarity, Solutions, Dilutions
- Topic: Homework problem G5
- Replies: 2
- Views: 629
Re: Homework problem G5
Ok, so the molarity of the sodium carbonate is 0.0797 M. For part a, they have given you 2.15 mmol of Na+. That means, you have the number of moles and you have the molarity of the compound. So, you would just use the molarity formula: M (molarity) = n(number of moles) / V. You can plug in the numbe...
- Wed Oct 10, 2018 5:59 pm
- Forum: Empirical & Molecular Formulas
- Topic: Question M19: Empirical and Molecular formula of caffeine [ENDORSED]
- Replies: 4
- Views: 1593
Re: Question M19: Empirical and Molecular formula of caffeine [ENDORSED]
As you said, when something is combusted, it reacts with oxygen to produce CO2 and water, H2O, and you aren't sure what elements are in caffeine. But, if you write the chemical equation out: Caffeine (unknown formula) + O2 -> CO2 + H2O + N2. Looking at the chemical equation, the reactants are caffei...
- Mon Oct 08, 2018 10:40 pm
- Forum: Limiting Reactant Calculations
- Topic: Fundamentals Problem M.11
- Replies: 2
- Views: 348
Re: Fundamentals Problem M.11
So, you know that in the second reaction, O2 is the limiting reactant. That means in the second reaction, the excess reactant (the one that there is too much of left) is going to P4O6 and they are asking us how much of that will be left. Remember that in reaction one (the first reaction), the produc...
- Mon Oct 01, 2018 5:07 pm
- Forum: Balancing Chemical Reactions
- Topic: Need help w/ determining the moles of gas produced
- Replies: 2
- Views: 202
Re: Need help w/ determining the moles of gas produced
The chemical equation is: 4 C4H10 (g) + 26 O2 (g) -> 16 CO2 (g) + 20 H2O (g) You can find the net number of moles produced by looking at the stoichiometric coefficients. You take the sum of coefficient of product and minus sum of coefficient of reactants to get the net number of moles produced. 4+ 2...
- Mon Oct 01, 2018 4:32 pm
- Forum: Limiting Reactant Calculations
- Topic: Limiting Reactant
- Replies: 2
- Views: 1005
Re: Limiting Reactant
The reason he says that you multiply 1.56 (moles of CaC2) by 2 is because in the balanced chemical equation, it says that 1 mole of CaC2 reacts with 2 mole of H2O. 1 CaC2 (s) + 2 H2O (l) -> Ca(OH)2 (aq) + C2H2 (g) By looking at the stoichiometric coefficients in the chemical equation, we see that fo...
- Mon Oct 01, 2018 3:18 pm
- Forum: Limiting Reactant Calculations
- Topic: Audio Visual Question 22
- Replies: 3
- Views: 467
Re: Audio Visual Question 22
Step 1: Check if the chemical equation is balanced. In this case, it isn't. Thus, the rewritten, correct chemical equation would be: C6H9Cl3 + 3AgNO3 -> 3 AgCl +C6H9(NO3)3. Then, it always helps if you write out what they give you, what they do not give you, and what you are trying to find. In this ...