Search found 102 matches
- Wed Mar 10, 2021 9:40 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: pt in cell diagram
- Replies: 2
- Views: 168
Re: pt in cell diagram
Just to add on, Pt is one of the possible electrodes that can be used when there is no conducting solid in a reaction. If a reaction has only ions involved, Pt can serve as the electrode for that half reaction. Pt is often used as a conducting solid but there are also other conducting solids that ca...
- Wed Mar 10, 2021 9:35 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Cell Reduction Potential
- Replies: 2
- Views: 140
Re: Cell Reduction Potential
I believe that electrolytic cells are when the overall cell potential is negative so some current must be applied to make the reaction occur. The overall cell potential can be negative and yes, this would mean the cell is not spontaneous without an outside input of current.
- Wed Mar 10, 2021 9:31 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: electrode
- Replies: 3
- Views: 247
Re: electrode
In this case, Pt is only in contact with the right side of the cell because that half reaction does not have a conducting solid that would allow electrons to flow through. On the left side, Cu serves as the conducting solid so Pt is not needed as an electrode. There are two electrodes in this cell, ...
- Tue Mar 09, 2021 3:33 pm
- Forum: Balancing Redox Reactions
- Topic: strongest reducing agent and strongest oxidizing agent
- Replies: 2
- Views: 186
Re: strongest reducing agent and strongest oxidizing agent
Yes, the best reducing agents have lower standard reduction potentials and the best oxidizing agents have higher standard reduction potentials. This is because a reducing agent wants to be oxidized so it would have a low reduction potential, and opposite for oxidizing agents.
- Mon Mar 08, 2021 2:27 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Writing Balanced reactions from Galvanic Cells
- Replies: 2
- Views: 189
Re: Writing Balanced reactions from Galvanic Cells
Sometimes the textbook starts off by writing both half reactions as reductions then flipping one so that the equations are the correct orientation. As long as you have the correct half reactions in the end and correct order of overall reaction, you should be fine. You just need to make sure that you...
- Fri Mar 05, 2021 9:12 am
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: 6L.1
- Replies: 4
- Views: 307
Re: 6L.1
The value for n is the number of moles of electrons transferred. You can find this value by looking at the half reactions, balancing them so the electrons cancel out in the overall reaction, and using the value in the balanced half reactions. Using this value should give you the correct answer. Hope...
- Thu Mar 04, 2021 9:22 pm
- Forum: Zero Order Reactions
- Topic: Saturated
- Replies: 3
- Views: 291
Re: Saturated
Just to add on, an enzyme or catalyst is saturated when they are all currently filled with substrate/reactant so the reaction rate is at its max. Adding more reactant would not increase the rate of reaction since the enzymes or catalysts are already working the fastest they can to turn reactant into...
- Wed Mar 03, 2021 1:55 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: When is E =0 and G=0 in Concentration Cell?
- Replies: 1
- Views: 146
Re: When is E =0 and G=0 in Concentration Cell?
Yes, when E naught and \Delta G naught are 0 it is referring to everything in standard form. I think you may be referring to \Delta G instead of G, because we mainly deal with \Delta G . Standard form is when the liquids and gases are pure, it is 1 atm, and solutions are 1 M at a specified temperatu...
- Tue Mar 02, 2021 9:02 am
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Nernst Equation
- Replies: 2
- Views: 201
Re: Nernst Equation
By this he means that the concentrations of the reactants and products at a point in time in the Galvanic cell influences the cell potential. The composition of the Galvanic cell influences the cell potential because of the lnQ term, where different values of Q at different times influences the valu...
- Mon Mar 01, 2021 11:42 pm
- Forum: Balancing Redox Reactions
- Topic: balancing redox reaction is basic solutions
- Replies: 8
- Views: 592
Re: balancing redox reaction is basic solutions
For balancing basic solutions, instead of adding H+ to the side that needs more H+, you add H2O to the side that needs more H+ and add OH- to the opposite side. The effect of adding 1 H2O to one side and an OH- to the other side is you have added a net 1H+ ion to one side of the equation to help bal...
- Fri Feb 26, 2021 12:49 pm
- Forum: Balancing Redox Reactions
- Topic: Oxidizing Vs Reducing agent
- Replies: 39
- Views: 1884
Re: Oxidizing Vs Reducing agent
Just to add on, you can think about how the oxidizing agent (which is the reactant being reduced) is what takes electrons from the reactant being oxidized, allowing this oxidation to occur. The reducing agent (the reactant being oxidized) allows the reduction to occur because it provides electrons t...
- Thu Feb 25, 2021 3:46 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Cell Diagram
- Replies: 3
- Views: 216
Re: Cell Diagram
Platinum is used when the half reaction does not have a solid form. For example, with Fe2+ and Fe3+, both are ions so they would be dissolved in solution, not solid. You would need to use Platinum for the reaction to be able to occur and electrons to be able to flow.
- Wed Feb 24, 2021 8:16 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Eºcell
- Replies: 11
- Views: 619
Re: Eºcell
If the reaction is under standard conditions and the Eº of the cell is positive, the reaction is spontaneous. However, the Eº of the cell can be positive but if the cell is not under standard conditions, such as a really high concentration of the products, then the reaction may not be spontaneous, e...
- Tue Feb 23, 2021 3:49 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: E cell vs Ecell naught
- Replies: 25
- Views: 1578
Re: E cell vs Ecell naught
Yes, as mentioned above, E cell naught is under standard conditions. This means that the E cell naught value does not change as the reaction proceeds, even though the E cell value does change.
- Tue Feb 23, 2021 3:47 pm
- Forum: Balancing Redox Reactions
- Topic: flipping reactions
- Replies: 5
- Views: 287
Re: flipping reactions
You should be given the overall equation or the cell diagram and you can look at this to determine which reactant is being reduced and which is being oxidized. If in the overall equation you see one reactant is losing electrons, you know it's being oxidized and so the half reaction would be that rea...
- Wed Feb 17, 2021 9:57 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Units for Temperature
- Replies: 5
- Views: 311
Re: Units for Temperature
I think most of the equations we use in thermodynamics use Kelvin. One case I know that Celsius is often used is in the q=mc\Delta T equation where often the C value and the temperature have units of Celsius in them. I would always just check the units to make sure everything cancels correctly. Hope...
- Wed Feb 17, 2021 9:52 am
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Which R Values to use
- Replies: 22
- Views: 8095
Re: Which R Values to use
When I want to know which R value to use, I look at the units that are in my formula already, look at the units I want my answer to be in, then determine which units need to be in the R value so the units will cancel correctly. If you compare units and find that the answer has the correct units, you...
- Tue Feb 16, 2021 9:02 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Cv and Cp values for an ideal gas
- Replies: 4
- Views: 942
Cv and Cp values for an ideal gas
What are we supposed to know about calculating the Cv and Cp values for an ideal gas in the equation delta S = nC ln(T2/T1). I know on the equation sheet it says that Cp = (5/2)R and Cv = (3/2)R but what does this mean and how is this derived? Also when will the equation U = 3/2 nRT (for an ideal ga...
- Tue Feb 16, 2021 6:29 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Delta H and Delta S both positive
- Replies: 31
- Views: 8084
Re: Delta H and Delta S both positive
One way to think of this is if you are subtracting a positive value from a positive value, delta G would only be negative and therefore spontaneous when the second term is bigger than the first. Only when the temperature is high enough is the reaction spontaneous.
- Tue Feb 16, 2021 8:47 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: What does the d in dq mean
- Replies: 3
- Views: 341
Re: What does the d in dq mean
The dq means a very small change in q. If you add up many small changes, then you can get a larger change in q. This notation of dq is used in integrals in order to show the change in q from one value to another, in a series of very small steps.
- Fri Feb 12, 2021 10:22 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Spontaneous as Written
- Replies: 10
- Views: 503
Re: Spontaneous as Written
If the delta G of a reaction is negative, then that reaction will be spontaneous under those conditions. I'm not sure what you mean by spontaneous as written, but I hope this answers your question!
- Fri Feb 12, 2021 10:18 am
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: HW 4A.7
- Replies: 4
- Views: 191
Re: HW 4A.7
For part b, you can take the amount of heat supplied to heat up the water divided by the total heat to heat up both the water and the copper kettle. For part a you have to do q = mc \Delta T twice, once for the copper kettle and once for the water. To find the percentage of heat used to heat up the ...
- Thu Feb 11, 2021 9:49 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Delta G and Spontaneity
- Replies: 10
- Views: 744
Re: Delta G and Spontaneity
Like they said above, a reaction is spontaneous if delta G is negative. In class he related this to delta S(univ) by adjusting the \Delta G = \Delta H-T\Delta S equation. He divided both sides by -T, simplified, and used his result to show that for delta S(univ) to be positive, delta G must be negat...
- Tue Feb 09, 2021 12:58 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Textbook 4.5
- Replies: 1
- Views: 111
Re: Textbook 4.5
The textbook assumes that it took 10 hours for the ice to melt. Because the water at 0 degrees Celsius took 0.5 hours to get to 5 degrees Celsius, and the ice took 10.5 hours to melt and reach 5 degrees Celsius, you can assume that 0.5 of those hours was the the fully melted ice being raised to 5 de...
- Mon Feb 08, 2021 7:11 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: delta q
- Replies: 3
- Views: 244
Re: delta q
At a constant volume, q = ∆U because w = 0. We know that ∆U = q + w and at a constant volume no expansion work is being done so w = 0 and ∆U = q. At a constant pressure q = ∆H because we know that heat released or absorbed at a constant pressure is equal to enthalpy. If the system is at a constant p...
- Thu Feb 04, 2021 9:21 pm
- Forum: Calculating Work of Expansion
- Topic: 4A3C
- Replies: 2
- Views: 106
Re: 4A3C
I think the answer key in the textbook is wrong since it says 8J but if you go to the answer key with the full explanation it says w = 28J. I believe that 28J should be the correct answer, and you can find this under resources on the main sapling page.
- Thu Feb 04, 2021 5:04 pm
- Forum: Calculating Work of Expansion
- Topic: Work notation
- Replies: 10
- Views: 400
Re: Work notation
I think you are supposed to use lowercase w for work since uppercase W is used to represent degeneracy.
- Thu Feb 04, 2021 9:40 am
- Forum: Calculating Work of Expansion
- Topic: integral equation
- Replies: 7
- Views: 295
Re: integral equation
The integral equation would be used when the external pressure is changing, so this would be when a reversible expansion is occurring and a change in volume of the system results. The equation w = -P(delta V) can be used when the external pressure is constant, since P will be one value that can be p...
- Wed Feb 03, 2021 4:25 pm
- Forum: Calculating Work of Expansion
- Topic: Spontaneous
- Replies: 26
- Views: 1450
Re: Spontaneous
A spontaneous reaction will occur under the current conditions without changing anything. As stated above, you must take into account not only enthalpy but also entropy to determine this, so an exothermic reaction is not always spontaneous. I think Dr. Lavelle will explain more in future lectures ab...
- Tue Feb 02, 2021 8:23 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Textbook Problem 4A.13
- Replies: 2
- Views: 94
Re: Textbook Problem 4A.13
Just add on, the q value will be negative for a system if that system is losing heat to the surroundings. If the system is gaining heat, then the q would be positive. It's the opposite when talking about the surroundings, since whatever a system gains the surroundings lose, and vice versa. So if the...
- Fri Jan 29, 2021 6:41 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Calculating standard enthalpy of formation clarification
- Replies: 7
- Views: 330
Re: Calculating standard enthalpy of formation clarification
I think for Hess's law, you are given the equations and their delta H values and you just have to find the way to combine them that gets you a desired overall equation. For standard enthalpy of formation, these are given in a table usually and you can use them to determine the standard enthalpy of a...
- Fri Jan 29, 2021 6:32 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Textbook Problem 6E.1
- Replies: 2
- Views: 109
Re: Textbook Problem 6E.1
For this problem, H2SO4 is a strong polyprotic acid so you have to account for both deprotonations. The first deprotonation is complete because the acid is strong, so it produces 0.15M HSO4-, and 0.15M H3O+. The next step is to use the Ka2 for the second deprotonation of the acid, which involves HSO...
- Wed Jan 27, 2021 4:24 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Textbook 5J5 (d)
- Replies: 2
- Views: 100
Re: Textbook 5J5 (d)
I think the equation for this problem is wrong in the book it's supposed to be 2HD(g) -> H2(g) + D2(g), so there are the same amount of moles on each side.
- Tue Jan 26, 2021 5:11 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Sapling #5 Reverse Equation
- Replies: 2
- Views: 144
Re: Sapling #5 Reverse Equation
Yes, the reason you need to flip equation 4 is because the product in the overall reaction is MCl3(s), not MCl3(aq). Flipping equation 4 would allow the final equation to have the correct state of MCl3.
- Tue Jan 26, 2021 10:56 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Predicting how compression favors reactants vs products
- Replies: 1
- Views: 107
Re: Predicting how compression favors reactants vs products
If there are more moles of gas on the product side, the reactants will be favored. If there are more moles of gas on the reactant side, the products will be favored. If there are the same amount of moles of gas on both sides, neither side is favored over the other. This is a shortcut to figure out w...
- Wed Jan 20, 2021 10:52 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Question from lecture #6
- Replies: 4
- Views: 246
Re: Question from lecture #6
Just to clarify, it's the H3O+ concentration you calculated from the ice table at equilibrium which was produced by the acid that must be less than 10^-7. If you calculate the H3O+ concentration to be less than 10^-7, you must account for the autoprotolysis of water which creates 10^-7 M of H3O+ and...
- Tue Jan 19, 2021 10:04 pm
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: Calculating the pH and percentage deprotonation of a weak acid
- Replies: 8
- Views: 477
Re: Calculating the pH and percentage deprotonation of a weak acid
Yes, the K value must be 10^-4 or smaller, but another way to check to make sure the approximation is valid is to check the percent ionization of the acid or base. If the value of x divided by the initial concentration of acid or base is less than 5%, then the approximation is valid. If the K value ...
- Mon Jan 18, 2021 10:49 am
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: neutral solution
- Replies: 9
- Views: 655
Re: neutral solution
Just to add, by this he means that if you calculate the [H3O+] to be less than 10^-7, then the amount of H3O+ normally in water caused by the autoprotolysis reaction influences the pH more than the amount of H3O+ generated by an acid. If you calculate that the acid generates 10^-9 M of H3O+ ions, th...
- Mon Jan 18, 2021 10:37 am
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: pH for Strong Acids
- Replies: 3
- Views: 423
Re: pH for Strong Acids
The pH for strong acids is just the negative log of the molar concentration because for strong acids, all of the acid dissociates into H3O+ ions and A- ions (A- refers to the conjugate base of an acid). If the strong acid completes dissociates into H3O+ and A-, then the initial concentration of the ...
- Mon Jan 18, 2021 10:32 am
- Forum: Phase Changes & Related Calculations
- Topic: ionization percentage
- Replies: 10
- Views: 418
Re: ionization percentage
Ionization percentage is the amount of acid or base that has turned into its conjugate acid or base. The formula for ionization percentage for acids is [A-]/[HA] * 100% and for bases it's [HB+]/[B] * 100%. The value of HA and B are the initial concentrations of the acid and base before equilibrium h...
- Fri Jan 15, 2021 6:27 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Hw promblem 6.3
- Replies: 5
- Views: 153
Re: Hw promblem 6.3
Because the initial solution was supposed to be a concentration of 0.025 M, and it is a strong acid, the concentration of H+ would have been 0.025 M. You can find the desired pH by doing -log[0.025]. To find the actual pH, you can use the formula M1V1 = M2V2 to solve for M2. M1 would be 0.025M, V1 ...
- Fri Jan 15, 2021 1:44 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Hw promblem 6.3
- Replies: 5
- Views: 153
Re: Hw promblem 6.3
Because the initial solution was supposed to be a concentration of 0.025 M, and it is a strong acid, the concentration of H+ would have been 0.025 M. You can find the desired pH by doing -log[0.025]. To find the actual pH, you can use the formula M1V1 = M2V2 to solve for M2. M1 would be 0.025M, V1 w...
- Thu Jan 14, 2021 1:29 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Gas
- Replies: 16
- Views: 574
Re: Gas
I think you would be given the units of the pressure of a gas. If they ask for a different unit in the answer, then you could just convert between the units, but bar and atm are very similar in value so there is not much difference between their values.
- Wed Jan 13, 2021 7:42 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Net Ionic Equation
- Replies: 3
- Views: 110
Re: Net Ionic Equation
In this problem, anything that is aqueous on both sides of the equation would not be in the net ionic equation since it does not change. In this case, Na and NO3 would be removed to form the net ionic equation because they are aqueous on both sides of the equation. Then you just need to take what is...
- Tue Jan 12, 2021 2:58 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Sapling Focus Question 5.61
- Replies: 3
- Views: 138
Re: Sapling Focus Question 5.61
I believe that adding water to the reaction would not affect the equilibrium since water is a liquid so it would not be included in the equilibrium expression. Because water is in excess and the concentration does not change very much before or after the reaction, adding water to the reaction does n...
- Tue Jan 12, 2021 2:52 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Textbook Focus Problem 5.33 a
- Replies: 3
- Views: 169
Re: Textbook Focus Problem 5.33 a
You can tell this reaction would be endothermic because it involves breaking the bond between the two atoms of X. Anytime you are breaking bonds it requires energy so the reaction would be endothermic, taking in heat in order to power the reaction.
- Fri Jan 08, 2021 5:45 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Self Test 5G.3: Using a Net Ionic Equation to Write an Equilibrium Constant
- Replies: 1
- Views: 81
Re: Self Test 5G.3: Using a Net Ionic Equation to Write an Equilibrium Constant
For this problem, anything that is aqueous on both sides of the equation would not be in the net ionic equation since it does not change. So in this case Na and NO3 would be removed to form the net ionic equation. This would leave you with 2Ag+(aq) + 2OH-(aq) <--> Ag2O(s) + H2O(l).
- Fri Jan 08, 2021 2:03 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Sapling #3
- Replies: 5
- Views: 157
Re: Sapling #3
For this problem, you can start by making an ICE table with H2, I2, and HI. You start with 0.300 M of both H2 and I2 based on the moles and liters they give you in the problem. You start with 0 M of HI. You can put these values into the initial concentration row. For the change in concentration, H2 ...
- Thu Jan 07, 2021 8:57 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 5.i #11
- Replies: 4
- Views: 161
Re: 5.i #11
Before you plug the values into the equation, you must convert them into moles/L. They start by giving you the values in mmol and give you the volume of 0.500 L so you must convert the mmol to mol and then divide that value by 0.500 L for each. That should give you the correct answer.
- Thu Jan 07, 2021 7:02 pm
- Forum: Ideal Gases
- Topic: Partial Pressure Gas Question
- Replies: 4
- Views: 173
Re: Partial Pressure Gas Question
Based on the problem and what you said you did, you should get the correct answer. The equation to use would be PV = nRT or P = nRT/V which is the same thing as P = conc(RT). If you plug in the concentration, correct R value, and temperature in Kelvin, you should get the correct answer. I would doub...
- Tue Jan 05, 2021 9:18 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Question #16 from Chemical Equilibrium Part 4 Video Module
- Replies: 2
- Views: 207
Re: Question #16 from Chemical Equilibrium Part 4 Video Module
Just to give another way to think about it, heating a reaction would favor the reaction in the endothermic direction. Since the forward reaction is exothermic, the endothermic reaction would be the reverse reaction so this would be favored. Therefore, the equilibrium would shift left. Hope this helps!
- Fri Dec 11, 2020 2:04 pm
- Forum: Properties & Structures of Inorganic & Organic Acids
- Topic: resonance and acid strength
- Replies: 3
- Views: 364
Re: resonance and acid strength
Something with more resonance would be a stronger acid because the electrons would be delocalized. Electron delocalization stabilizes molecules and therefore leads to a more stable conjugate base of an acid. If a molecule has more double bonds, there is resonance to delocalize electrons, so this mol...
- Thu Dec 10, 2020 8:49 pm
- Forum: Amphoteric Compounds
- Topic: Amphoteric Oxides
- Replies: 2
- Views: 601
Re: Amphoteric Oxides
I think the main way to tell is just that amphoteric oxides follow a diagonal line along the metalloids. The oxides with elements on the left side of the periodic table are basic and those with elements on the right side of the periodic table are acidic, so those in between are amphoteric. These amp...
- Wed Dec 09, 2020 6:22 pm
- Forum: Trends in The Periodic Table
- Topic: effective nuclear charge and shielding
- Replies: 4
- Views: 253
Re: effective nuclear charge and shielding
Just to add on, when going left to right across a period, the number of shielding electrons does not increase but the number of protons does increase. This means that the valence electrons are being pulled in by a greater force (from more protons) but there are not more shielding electrons to counte...
- Tue Dec 08, 2020 8:22 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Formal charge when doing VSEPR
- Replies: 2
- Views: 170
Re: Formal charge when doing VSEPR
In general you should use formal charges to determine Lewis structures to ensure you have the correct structure. However, when determining molecular shape, it does not matter if a bond is a single or double bond because it is counted as one region of electron density. So if you have a structure with...
- Tue Dec 08, 2020 10:49 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: textbook 2E17
- Replies: 2
- Views: 166
Re: textbook 2E17
To find when the bond angle is slightly less than a value, you look at the number of areas of high electron density. This would include bond areas and lone pairs. In O3, there are two bond regions and one lone pair region, meaning there are 3 regions of electron density. Three regions of electron de...
- Fri Dec 04, 2020 7:15 pm
- Forum: Naming
- Topic: Chelating Ligands
- Replies: 3
- Views: 229
Re: Chelating Ligands
Chelating ligands are ligands that attach to a metal atom in two or more areas. This would make them polydentates.
- Fri Dec 04, 2020 7:11 pm
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: Relative Acidity
- Replies: 2
- Views: 123
Re: Relative Acidity
An electron withdrawing atom would be an atom that has the power to pull some electrons away from the atom it is bonded to in order to stabilize the negative charge on that atom by delocalizing electrons. An example he gave is Cl bonded to O versus Br bonded to O in ClOH and BrOH. Cl has a higher el...
- Thu Dec 03, 2020 1:49 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Tetrahedral vs Square Planar
- Replies: 6
- Views: 301
Re: Tetrahedral vs Square Planar
I think we are just supposed to know that if the coordination number is 4, it could be square planar or tetrahedral, but we don't have to know how to tell which one it is.
- Wed Dec 02, 2020 8:36 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: [Pt(en)Cl2] coordination number
- Replies: 2
- Views: 392
Re: [Pt(en)Cl2] coordination number
[Pt(en)Cl2] would have a coordination number of 4. This is because there are in total 4 different bonds to the central atom, Pt. There are two Cl which each contribute one bond to the central atom. The (en) stands for ethylenediamine which is a bidentate, meaning it forms two bonds to the central at...
- Wed Dec 02, 2020 9:17 am
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Sapling Week 9 HW Question 2
- Replies: 11
- Views: 577
Re: Sapling Week 9 HW Question 2
The coordination number is the number of bonds to the transition metal atom in the coordination complex. The elements in the bracket are the only elements directly bonded to the transition metal atom. Fe would be the central transition metal atom and the 4 Br atoms would be bonded to it, so therefor...
- Mon Nov 30, 2020 10:48 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: trigonal bipyramidal electron geometry
- Replies: 2
- Views: 311
Re: trigonal bipyramidal electron geometry
For the trigonal bipyramidal arrangement, if the lone pair was in the axial position, it would be 90 degrees from three different bonds (the bonds at the equatorial positions). However, if the lone pair was in the equatorial position, it would only be 90 degrees from 2 bonds (the bonds in the axial ...
- Sat Nov 28, 2020 11:46 am
- Forum: Polarisability of Anions, The Polarizing Power of Cations
- Topic: Polar vs. nonpolar bonds/molecules
- Replies: 8
- Views: 436
Re: Polar vs. nonpolar bonds/molecules
If there are two of the same atom on each side of a central atom (with no lone pairs on the central atom), the overall molecule would be nonpolar. However, the individual bonds inside the molecule are still polar bonds because they are between different atoms, but because the molecule is linear, it ...
- Fri Nov 27, 2020 8:31 pm
- Forum: Sigma & Pi Bonds
- Topic: Sapling Learning Week 7 and 8 Homework Question 16
- Replies: 4
- Views: 337
Re: Sapling Learning Week 7 and 8 Homework Question 16
Something has a delocalized pi bond when it has resonance structures that each have a different location for one pi bond. In other words, if a pi bond can be drawn in different places, then that pi bond would be delocalized. For this question, out of the three in the first part of the question that ...
- Wed Nov 25, 2020 12:39 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Negative Pole on Molecule
- Replies: 2
- Views: 105
Re: Negative Pole on Molecule
The negative pole on a molecule would be on the atom in a bond with a higher electronegativity. If the atoms in a bond are not the same, there will be a negative and positive pole, even if it is very small. If oxygen is bonded to hydrogen, oxygen would have the negative pole because it has a higher ...
- Tue Nov 24, 2020 10:26 pm
- Forum: Ionic & Covalent Bonds
- Topic: Valence in d-block
- Replies: 4
- Views: 286
Re: Valence in d-block
To find the valence electrons in the d-block, you should include the s-block electrons from the period of that element as well as any d-block electrons in that period. You can do this by counting over from the left side of the periodic table towards the element. For example, Mn would have 7 valence ...
- Tue Nov 24, 2020 2:00 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Shape vs. Electron Arrangement
- Replies: 4
- Views: 243
Re: Shape vs. Electron Arrangement
Often times, the electron arrangement differs from the shape when there are lone pairs in a molecule. The electron arrangement of a molecule is where there are regions of electron density, whether that be bonds or lone pairs. However, when finding the shape of a molecule, you only pay attention to t...
- Fri Nov 20, 2020 11:05 am
- Forum: Resonance Structures
- Topic: Sapling Resonance Structures
- Replies: 4
- Views: 333
Re: Sapling Resonance Structures
Another thing to note is that if a resonance structure is very unfavorable, it will contribute very little to the overall resonance hybrid, but it still contributes a very small amount because it is a possible resonance structure. The most favorable and lowest energy resonance structures contribute ...
- Thu Nov 19, 2020 9:18 am
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Textbook Exercise 2A.17
- Replies: 1
- Views: 136
Re: Textbook Exercise 2A.17
Mn would normally have 5 electrons in the 3d subshell and 2 electrons in the 4s subshell. If it were to lose 3 electrons, it would lose them from the 4s subshell first because it is higher in energy when it is filled and then it would lose 2 additional electrons from the 3d subshell to lose a total ...
- Wed Nov 18, 2020 1:26 pm
- Forum: Lewis Structures
- Topic: Lewis Structure Covalent Bonds
- Replies: 2
- Views: 108
Re: Lewis Structure Covalent Bonds
Often Lewis structures are used to show covalent bonds, but there are also Lewis structures to represent ionic bonds. An example would be for NaCl, where you would draw a Na+ ion with a bracket showing its + charge and a Cl- ion next to it with 8 lone pairs representing its full valence shell, as we...
- Wed Nov 18, 2020 1:16 pm
- Forum: Resonance Structures
- Topic: Resonance
- Replies: 11
- Views: 637
Re: Resonance
Just to add, elements want to have the same amount of electrons belonging to them as they have valence electrons, so the more atoms that can fulfill this requirement the better the structure. Some atoms are more likely to accept a negative formal charge due to their electronegativity, and some atoms...
- Mon Nov 16, 2020 11:56 pm
- Forum: Dipole Moments
- Topic: Dipole Moments Always Exist?
- Replies: 3
- Views: 178
Re: Dipole Moments Always Exist?
There are dipoles between any two atoms that aren't the same, they are just very small for some atoms. Between C and H, there is a dipole but it is very small because their electronegativities are very similar. If the same atom is bonded to itself, there is no dipole. Also one thing to note is that ...
- Wed Nov 11, 2020 2:04 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: 4s and 3d Orbitals
- Replies: 1
- Views: 99
Re: 4s and 3d Orbitals
The 4s orbital is not higher in energy until it has been filled. So for K, Ca, and any element after Ca, the 4s is lower in energy while it is empty. This would cause electrons to fill that subshell before the 3d subshell. Once the 4s subshell is filled, it would become higher in energy than 3d agai...
- Wed Nov 11, 2020 2:01 pm
- Forum: Bond Lengths & Energies
- Topic: Energy per mole
- Replies: 2
- Views: 98
Re: Energy per mole
Yes, I believe that for 1 mole of Na+ and 1 mole of Cl- interacting, the total energy of their interactions is -250 kJ.
- Mon Nov 09, 2020 8:45 pm
- Forum: Ionic & Covalent Bonds
- Topic: Textbook Question 2A.15 Part D
- Replies: 1
- Views: 106
Re: Textbook Question 2A.15 Part D
The reason Ga would form an ion of 3+ is because it has 3 electrons in its outer shell, 1 in the 4p subshell and 2 in the 4s subshell. If Ga were to become an ion, it would lose all of the electrons in its outermost shell, which would be 3 electrons. It would not lose the electrons in the 3d subshel...
- Mon Nov 09, 2020 8:41 pm
- Forum: Resonance Structures
- Topic: oxidation numbers in regards to resonance
- Replies: 2
- Views: 97
Re: oxidation numbers in regards to resonance
For that question, I believe the way to solve it is to find the atom with the formal charge as close as possible to the oxidation number. So if the oxidation number of an atom is 7+, you would choose the resonance structure with the atom that has a formal charge closest to 7+. Hope this helps!
- Mon Nov 09, 2020 6:31 pm
- Forum: Formal Charge and Oxidation Numbers
- Topic: Lowering Formal Charges
- Replies: 7
- Views: 372
Re: Lowering Formal Charges
For some molecules, not all the formal charges can be zero. This is because if you add up all the formal charges of atoms in a molecule, you get the overall charge of that molecule. So for a molecule with a 2+ charge, the formal charges of the atoms have to add up to that value so some of the atoms ...
- Fri Nov 06, 2020 9:03 am
- Forum: Resonance Structures
- Topic: Ample and overwhelming bond character
- Replies: 3
- Views: 548
Re: Ample and overwhelming bond character
If you are referring to the sapling homework, they are talking about bond character in terms of the length of the bond between two atoms. If the bond is a double bond, that bond would be shorter than a single bond between the two same atoms. When sapling asks if a bond is ample in one property, it i...
- Thu Nov 05, 2020 11:56 pm
- Forum: Ionic & Covalent Bonds
- Topic: 2A.19 Ground state electron configuration for ions
- Replies: 3
- Views: 8688
Re: 2A.19 Ground state electron configuration for ions
For Ni(2+), the Nickel neutral atom has lost 2 electrons. The electron configuration for a neutral Nickel atom would be [Ar]3d84s2. When Nickel loses 2 electrons to become Ni(2+), the electrons are lost from the 4s subshell because when it is filled it is higher in energy than the 3d subshell. Becau...
- Wed Nov 04, 2020 10:19 am
- Forum: Lewis Structures
- Topic: Electrons in Lewis Structures
- Replies: 8
- Views: 352
Re: Electrons in Lewis Structures
To determine the valence electrons for each element, you can look at the group number of that element. For group 1 and 2, they each have 1 and 2 valence electrons respectively, meaning they have the same amount of valence electrons as their group number. For groups 13-18, you have to subtract 10 fro...
- Wed Nov 04, 2020 9:39 am
- Forum: Quantum Numbers and The H-Atom
- Topic: subshells and orbitals
- Replies: 9
- Views: 2094
Re: subshells and orbitals
To expand, orbitals are part of subshells, and can hold either the same or less electrons. One example of a subshell would be 2s, which can only hold 2 electrons because it contains one orbital (and each orbital holds 2 electrons). However, the subshell 2p has three orbitals in it so it can hold 6 e...
- Tue Nov 03, 2020 5:12 pm
- Forum: Trends in The Periodic Table
- Topic: Ionic Radius
- Replies: 8
- Views: 1566
Re: Ionic Radius
For elements with the same number of electrons, the atom with more protons in the nucleus or the higher effective nuclear charge has the smaller atomic radius. In the case of Na+ and F-, Na+ has the smaller ionic radius because it has a stronger pull from the nucleus, causing electrons to be pulled ...
- Fri Oct 30, 2020 1:59 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Sapling #24
- Replies: 5
- Views: 208
Re: Sapling #24
If you look at the periodic table, you can see that oxygen has 4 electrons in the p-block. The p-block only has 3 orbitals, so two of those electrons are paired up in one orbital with opposite spins. If you were to take one of those electrons away, there would be 3 p-block electrons and they could e...
- Thu Oct 29, 2020 3:28 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: l=4
- Replies: 13
- Views: 490
Re: l=4
I think he said in the lecture that in theory l could be equal to 4, but we don't have elements that are this high. So in the periodic table, there are no elements with l=4.
- Thu Oct 29, 2020 1:51 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Sapling Bohr orbits of the hydrogen atom
- Replies: 2
- Views: 101
Re: Sapling Bohr orbits of the hydrogen atom
Yes, the wave must connect at both ends by having a full wavelength. It must also have a constant wavelength and amplitude throughout the whole wave so it does not vary.
- Tue Oct 27, 2020 1:50 pm
- Forum: Photoelectric Effect
- Topic: Energy of photon
- Replies: 5
- Views: 202
Re: Energy of photon
I believe the energy of Ep is usually given in J/photon, but pay attention to the question to make sure your units match for each of the values in that equation. If you have the energy in J/photon, you can multiply by Avogadro's number to get J/mol of photon.
- Tue Oct 27, 2020 1:46 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: Textbook Problem 1B27
- Replies: 6
- Views: 409
Re: Textbook Problem 1B27
SI units are always in kg, not g. Because of this, the constant h used in the Heisenberg's indeterminacy equation is in units of J/s, and 1 J = 1 kg m^2 s^-2. Because the constant h uses kg, you must use units of kg for the mass.
- Fri Oct 23, 2020 8:22 am
- Forum: Limiting Reactant Calculations
- Topic: Ammonia Percent Yield (Workshop)
- Replies: 1
- Views: 225
Re: Ammonia Percent Yield (Workshop)
Yes, that would be the correct way to solve the problem. I got a slightly different answer for the yield of NO from 15.6g NH3 (I got 27.49g NO produced) but either way O2 is still the limiting reactant and your answer would be correct.
- Thu Oct 22, 2020 3:24 pm
- Forum: Photoelectric Effect
- Topic: Textbook Problem 1A. 9
- Replies: 4
- Views: 261
Re: Textbook Problem 1A. 9
For this problem, you can figure out the event based on the wavelength of the electromagnetic radiation. For your example, the wavelength is 340 nm, which is in the ultraviolet portion of the electromagnetic spectrum, so this event is acquiring a suntan. By comparing each wavelength calculated to th...
- Wed Oct 21, 2020 4:21 pm
- Forum: Properties of Electrons
- Topic: Properties of Electrons and Matter as a Whole
- Replies: 5
- Views: 242
Re: Properties of Electrons and Matter as a Whole
Yes, you are correct. While all matter has wavelike properties, it can only be detected at a small mass due to De Broglie's equation, wavelength = h/p. Also, electrons do have both wave and particle like properties.
- Tue Oct 20, 2020 4:05 pm
- Forum: Significant Figures
- Topic: How would you round these numbers
- Replies: 10
- Views: 620
Re: How would you round these numbers
I believe that rounding those answers to 8.0 and 5.0 is correct if using two sig figs.
- Mon Oct 19, 2020 4:23 pm
- Forum: Photoelectric Effect
- Topic: When to use E = h(nu) and not to
- Replies: 3
- Views: 173
Re: When to use E = h(nu) and not to
Yes, you are correct. Because E(photon) - work function = Ek, you could rearrange this so work function = E(photon) - Ek. This means that the energy required to remove an electron is the (energy of a photon) - (kinetic energy of the ejected electron).
- Wed Oct 14, 2020 8:12 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Series
- Replies: 2
- Views: 223
Re: Series
What those statements mean is that for each of the series, it involves electrons starting at a certain electron energy level and being excited to a higher level, or starting at various high energy levels and jumping back down to one standard energy level. For the Lyman series, which is in the ultrav...
- Wed Oct 14, 2020 7:55 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Wavelength for Atomic Spectroscopy
- Replies: 3
- Views: 208
Re: Wavelength for Atomic Spectroscopy
Yes, more energy is required for an electron to jump from n=1 to n=5 than from n=2 to n=4. Because of this, a shorter wavelength of light must be emitted when going from n=5 to n=1 because shorter wavelengths contain more energy. The transition from n=4 to n=2 would emit longer wavelength light beca...
- Tue Oct 13, 2020 9:24 pm
- Forum: Properties of Light
- Topic: Finding kinetic energy
- Replies: 3
- Views: 201
Re: Finding kinetic energy
You use the mass of an electron, 9.109 * 10^-31 kg, which can be found in the back of the book. You know to use this mass because you use the mass of the object you need to find the kinetic energy of. Since in this situation you are finding the kinetic energy of an electron, you use the mass of an e...
- Tue Oct 13, 2020 11:47 am
- Forum: Photoelectric Effect
- Topic: Light Intensity
- Replies: 5
- Views: 245
Re: Light Intensity
My understanding is that light intensity is simply how bright a light is, and increasing the intensity increases the amount of photons being emitted by that light source. The wavelength and frequency are independent of the amplitude, and are instead dependent on the length of a wave. So changing the...
- Tue Oct 13, 2020 11:33 am
- Forum: Quantum Numbers and The H-Atom
- Topic: Atomic Spectra Post Assessment #42
- Replies: 2
- Views: 103
Re: Atomic Spectra Post Assessment #42
For this problem, you start by turning the frequency into energy using E = hv. Then, you calculate the energy of the n=4 level by using the En = -hR/n^2 equation. Because Efinal - Einitial = difference in energy, you can plug in the energy difference you calculated using the frequency (which should ...
- Wed Oct 07, 2020 10:29 am
- Forum: Limiting Reactant Calculations
- Topic: Limiting reactants
- Replies: 8
- Views: 429
Re: Limiting reactants
If the molar ratio was equal to the calculated moles that would mean there is no limiting reactant because each reactant has the perfect amount of moles to react with each other. To find the amount of product formed you could use moles of either reactant because they both are able to produce the sam...
- Wed Oct 07, 2020 10:23 am
- Forum: Molarity, Solutions, Dilutions
- Topic: G.21 Homework Problem
- Replies: 7
- Views: 361
Re: G.21 Homework Problem
For this problem, I start by converting the 0.500 grams of KCl, K2S, and K3PO4 all into moles by using their molar masses. Once I had the moles, I knew the moles of KCl = moles of K, 1 mole of K2S = 2 mole K, and 1 mole K3PO4 = 3 mole K. I used these conversion factors to convert from moles of each ...
- Tue Oct 06, 2020 10:36 am
- Forum: Significant Figures
- Topic: Tips for counting sig figs?
- Replies: 9
- Views: 2410
Re: Tips for counting sig figs?
If a number has decimal places, it usually has the same amount of sig figs as digits in the number. For example, 4.00, 40.2, and 5.46 all have 3 sig figs because they all have 3 digits. However, in a number like 0.029, there are no digits present before the 2, so this value would only have 2 sig fig...