I'm not entirely sure about this, but I believe you can't tell whether the energy in each step increases or decreases. It can happen either way, depending on the individual delta H of each step.
Hope this helped!
Search found 110 matches
- Tue Mar 10, 2020 10:03 am
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: reaction profile
- Replies: 2
- Views: 201
- Tue Mar 10, 2020 10:00 am
- Forum: General Rate Laws
- Topic: 7a.15
- Replies: 2
- Views: 260
Re: 7a.15
The rate is independent of [C] because when you look at rates 1 and 4, you can see that while the concentration of C stays the same, the rates still change. As a result, C is zero'th order. Set the concentration of either A or B as constant and compare 2 rates. See how the rate changes as concentrat...
- Tue Mar 10, 2020 9:58 am
- Forum: Biological Examples
- Topic: adsorbtion
- Replies: 6
- Views: 806
Re: adsorbtion
To add on, adsorption is defined as reactants on the surface of the catalyst. For this to occur, I believe it is implied that the catalyst will have to be solid, so that there can be a surface for the reactants to rest on and react with. If it were a homogenous catalyst, the complex would likely for...
- Tue Mar 10, 2020 9:54 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 7B.3
- Replies: 2
- Views: 289
Re: 7B.3
An increase of 0.034 in B will result in a 0.068 decrease in A, because A decreases at twice the rate of B increasing. You then subtract 0.068 from [A]0 to get [A], which you can plug into your equation.
Hope this helped!
Hope this helped!
- Tue Mar 10, 2020 9:52 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 7B.9(a)
- Replies: 1
- Views: 170
Re: 7B.9(a)
Use the ratio between A and B to figure out how much A will decrease by when B increases. You can then figure out [A] by subtracting this value from [A]0.
Because B is 3 times the amount of A, A will decrease by 1/3 the rate of B. So, you divide B by 3.
Hope this helped!
Because B is 3 times the amount of A, A will decrease by 1/3 the rate of B. So, you divide B by 3.
Hope this helped!
- Mon Mar 02, 2020 10:45 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Different types of galvanic cells
- Replies: 2
- Views: 193
Re: Different types of galvanic cells
Electrolytic cells push the cell in a non-spontaneous direction. Therefore, Ecell will be negative, and ∆G will be positive.
Hope this helped!
Hope this helped!
- Mon Mar 02, 2020 10:40 am
- Forum: Interesting Applications: Rechargeable Batteries (Cell Phones, Notebooks, Cars), Fuel Cells (Space Shuttle), Photovoltaic Cells (Solar Panels), Electrolysis, Rust
- Topic: Metal in a Solution
- Replies: 3
- Views: 274
Re: Metal in a Solution
You have to look at the solubility constant, Ksp. This tells you how soluble a substance is, and whether it will dissolve in water.
Hope this helped!
Hope this helped!
- Mon Mar 02, 2020 10:36 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 6L 9
- Replies: 2
- Views: 197
Re: 6L 9
K+ and Cl- are spectator ions, and because this is a net ionic equation, you do not need to include them in the reaction.
When you are writing your half-reactions, only include the the species that are being oxidized or reduced.
Hope this helped!
When you are writing your half-reactions, only include the the species that are being oxidized or reduced.
Hope this helped!
- Mon Mar 02, 2020 10:29 am
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: 5J.13
- Replies: 5
- Views: 458
Re: 5J.13
You know that K decreases as temperature increases. This means that the reaction is exothermic. Because of this, the reactants are favored, and increasing the temperature will result in a shift to the reactant side. More ammonia will NOT be formed, because this is a product. You do not need an ICE ...
- Mon Mar 02, 2020 10:27 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Homework 6L.5 Part B and C
- Replies: 1
- Views: 154
Re: Homework 6L.5 Part B and C
I was confused about part B as well. I figured the I2 (s) would serve as one of the electrodes. For part C, however, HCl is included in the form of H+. In the solution, HCl will immediately dissociate to form H+ and Cl-. Because Cl- is a spectator ion, it will not be included in the oxidation half-r...
- Mon Feb 24, 2020 3:40 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Cell Diagrams
- Replies: 9
- Views: 572
Re: Cell Diagrams
Keep in mind that the species on the left of the double bar (the salt bridge) is not always the anode, and the species on the right of the double bar is not always the cathode. To determine which one is which, you have to determine the flow of the electrons by writing half reactions. Hope this helped!
- Mon Feb 24, 2020 3:33 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 6L.5 D
- Replies: 1
- Views: 123
Re: 6L.5 D
It's possible that you have to balance the equation, because you don't seem to have the same number of electrons transferred in both half-reactions. I would suggest writing the half reactions, including the electrons so it is easier to see the flow and how many are transferred.
Hope this helped!
Hope this helped!
- Mon Feb 24, 2020 3:31 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Gibbs Free Energy and K
- Replies: 5
- Views: 413
Re: Gibbs Free Energy and K
You can relate Gibbs free energy to K using the following equation:
∆Gº = -RTlnK
As you can see, this is not dependent on pressure or temperature. However, the max work that can be done is given by ∆G at constant pressure and temperature.
Hope this helped!
∆Gº = -RTlnK
As you can see, this is not dependent on pressure or temperature. However, the max work that can be done is given by ∆G at constant pressure and temperature.
Hope this helped!
- Mon Feb 24, 2020 3:27 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Flow of Galvanic cell
- Replies: 3
- Views: 198
Re: Flow of Galvanic cell
The cathode has a negative charge because of the electrons traveling toward it (remember, the electrons travel when delta G is positive, aka a NONspontaneous reaction – work is being done). Because of this, the cations will be attracted to the negative charge of the cathode and travel toward it. Hop...
- Mon Feb 24, 2020 3:22 pm
- Forum: Student Social/Study Group
- Topic: 2/24 lecture
- Replies: 5
- Views: 408
Re: 2/24 lecture
We also derived different forms of the Nernst equation and related the equation to K. An important takeaway from class was that when your system is at equilibrium, it is no longer doing work and your battery is dead. So, Ecell will be 0.
Hope this helped!
Hope this helped!
- Wed Feb 19, 2020 12:26 am
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Adiabatic systems
- Replies: 3
- Views: 309
Re: Adiabatic systems
Yes. q will always equal 0 for an adiabatic system because of the nature of the conditions. Heat is not allowed to transfer between the system and its surroundings, which is exactly what q measures! So, its value in adiabatic systems will always be zero.
Hope this helped!
Hope this helped!
- Wed Feb 19, 2020 12:24 am
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: 6L.1
- Replies: 1
- Views: 162
Re: 6L.1
Although I'm not too sure either, my best guess would be to use the deltaG = deltaG(naught) + RTlnQ equation. Solve for each part, relating Ecell to deltaG (I believe we will go over this in week 7 lectures).
Hope this helped!
Hope this helped!
- Wed Feb 19, 2020 12:21 am
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Relationship between free energy and cell potential?
- Replies: 2
- Views: 248
Re: Relationship between free energy and cell potential?
When G is negative, the reaction will occur spontaneously. However, when it is positive, energy must be put in for the reaction to occur. When G is 0, the reaction is at equilibrium (this is when your battery would be dead).
- Wed Feb 19, 2020 12:20 am
- Forum: Balancing Redox Reactions
- Topic: Acidic vs. Basic
- Replies: 1
- Views: 114
Re: Acidic vs. Basic
It's true that H+ ions make the solution acidic and more OH- ions make it basic. However, in the balanced reaction, I'm guessing you're confused about including H2O or not. In acidic solutions, we don't really include it – instead, we just show the dissociation. In basic solutions, we write out the ...
- Wed Feb 19, 2020 12:18 am
- Forum: Balancing Redox Reactions
- Topic: Oxidation Numbers
- Replies: 7
- Views: 422
Re: Oxidation Numbers
Halogens are always -1, oxygen is (almost) always -2, and hydrogen is always +1. In general, earth metals are +1 and alkaline earth metals are +2.
Hope this helped!
Hope this helped!
- Wed Feb 12, 2020 11:20 am
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: Different forms of entropy
- Replies: 2
- Views: 267
Re: Different forms of entropy
deltaS is referring to the entropy of the system. deltaS(total) refers to the entropy of the S(system) + S(surroundings).
Remember that S(total) = S(sys) + S(surr).
Hope this helped!
Remember that S(total) = S(sys) + S(surr).
Hope this helped!
- Wed Feb 12, 2020 11:19 am
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: midterm exam
- Replies: 1
- Views: 213
Re: midterm exam
4A through 4J will be on the midterm; however, there are some sections we do not need to know (e.g. the Born-Haber cycle). Also, we only need to know the basics of Gibbs free energy, as we haven't gone into enough detail on it in class.
Hope this helped!
Hope this helped!
- Wed Feb 12, 2020 11:17 am
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: External force
- Replies: 6
- Views: 419
Re: External force
There is always external force present, even in a reversible reaction. If there was no external force, there would be nothing to push against, and therefore no work being done. The thing to pay attention to is whether the external pressure is staying CONSTANT. With reversible pathways, the external ...
- Wed Feb 12, 2020 11:15 am
- Forum: Calculating Work of Expansion
- Topic: Determining Expansion Work
- Replies: 4
- Views: 618
Re: Determining Expansion Work
If there are more moles of gas on the product side than the reactant side, then the reaction did expansion work (expanded to more moles of gas). On the other hand, if there are fewer moles of gas on the product side, this means the system was compressed. When calculating the number of moles of gas o...
- Wed Feb 12, 2020 11:08 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: entropy at equilibrium
- Replies: 3
- Views: 241
Re: entropy at equilibrium
Entropy is a maximum at equilibrium because deltaS(system) = -deltaS(surroundings). At equilibrium, one side will gain all the possible entropy, while the other will lose it.
Hope this helped!
Hope this helped!
- Wed Feb 12, 2020 11:06 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 4f.1
- Replies: 2
- Views: 197
Re: 4f.1
Because entropy is a state function, it doesn't matter which pathway we take, because we will end up getting to the same end result. Reversible pathways are more convenient to calculate for, so we tend to use that. In reality, irreversible pathways will have a slightly lower entropy than reversible....
- Wed Feb 12, 2020 11:03 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Entropy in different phases
- Replies: 3
- Views: 187
Re: Entropy in different phases
Aqueous is more entropic than liquid because it is dissociated; however, it still has much less entropy than a gas at the same temperature.
Hope this helped!
Hope this helped!
- Wed Feb 12, 2020 11:02 am
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Examples of systems
- Replies: 3
- Views: 92
Re: Examples of systems
You have to read the conditions of the question to be able to tell if it is open, closed, or isolated. Evaluate what is allowed to be exchanged with the surroundings. Can energy be exchanged, but matter cannot? This would be a closed system.
Hope this helped!
Hope this helped!
- Wed Feb 12, 2020 10:59 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Reversible and irreversible
- Replies: 3
- Views: 241
Re: Reversible and irreversible
To add on, you can look at the graph in the textbook to get a better idea of the work being done in reversible v/s irreversible processes. In reversible processes, we are increasing the volume little by little and allowing the external pressure to adjust to equal the internal pressure. As a result, ...
- Mon Feb 03, 2020 11:13 am
- 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: Residual Entropy
- Replies: 3
- Views: 59
Re: Residual Entropy
Residual entropy is the difference between the entropy in a state that is NOT at equilibrium and the entropy of the most stable state of a substance close to absolute zero. I was a bit confused as well.
- Mon Feb 03, 2020 11:07 am
- Forum: Calculating Work of Expansion
- Topic: Calculus on The Midterm
- Replies: 8
- Views: 434
Re: Calculus on The Midterm
We aren't expected to go into this class with knowledge of calculus, so I don't think we will have to actually solve any integrals. Lavelle may, however, ask us to do a problem in a different way that will essentially give us the same answer an integral would – the area under a curve. Hope this help...
- Mon Feb 03, 2020 11:06 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Thermodynamic Systems
- Replies: 3
- Views: 165
Re: Thermodynamic Systems
There is an open system, a closed system, and an isolated system. In an open system, energy and matter can exchange between the system and the surroundings. In a closed system, matter cannot exchange, but energy can. In an isolated system, neither can exchange.
Hope this helped!
Hope this helped!
- Mon Feb 03, 2020 11:04 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: First Law
- Replies: 6
- Views: 143
Re: First Law
The first law of thermodynamics states that energy is never removed or destroyed, it is only transferred. This is helpful because we know that if energy (q) is leaving the system, it MUST be going into the surroundings. We can also use this law to make the statement that + q = - q. This essentially ...
- Mon Feb 03, 2020 11:03 am
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: q and w
- Replies: 10
- Views: 550
Re: q and w
Along with the tricks mentioned above, you can also determine whether work will be positive or negative based on expansion. If the system is expanding outward, work will be negative (w = -P(delta V)). On the other hand, if it is expanding inward, work will be positive.
Hope this helped!
Hope this helped!
- Mon Jan 27, 2020 9:04 pm
- Forum: Phase Changes & Related Calculations
- Topic: sig figs
- Replies: 6
- Views: 207
Re: sig figs
For questions with constants such as the 1 atm and 25 degrees Celsius, you do not consider these in determining sig figs. A good tip is to just follow the sig figs in the number given in the problem (molar amount, liters, etc). However, when calculating your answer, keep a lot of decimal places so y...
- Mon Jan 27, 2020 8:58 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Stirrer in Calorimeter?
- Replies: 3
- Views: 394
Re: Stirrer in Calorimeter?
The reaction will occur and eventually, we will have a homogenous reaction. However, in this time, heat may be lost to the surroundings. The stirrer allows us to mix the solution and make it homogenous faster so that the temperature is distributed evenly and our measurement may be more accurate. Hop...
- Mon Jan 27, 2020 8:55 pm
- Forum: Administrative Questions and Class Announcements
- Topic: 14A Final Answer Key
- Replies: 5
- Views: 204
Re: 14A Final Answer Key
I don't think there is an answer key for the final. Your best option might be to just search up the answers or compare with friends. It might also help to do the problem again and see if you get a different answer!
Hope this helped :)
Hope this helped :)
- Mon Jan 27, 2020 8:54 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Negative delta H
- Replies: 2
- Views: 102
Re: Negative delta H
You are correct in saying that bonds are breaking. However, you need to take into account the AMOUNT of energy needed to break these bonds. The reverse reaction would be negative because the energy needed to form the bonds in the products (releasing energy, negative G) may be greater than the energy...
- Mon Jan 27, 2020 8:51 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 4C.13
- Replies: 3
- Views: 106
Re: 4C.13
How did you calculate q for the ice? You should have added the q for the ice undergoing a phase change to the q for the ice melting to become 45 degrees. Then, you set this value to mC(delta t) of the 400.0g of water. Solve for the final temperature, given that the initial is 45 degrees Celsius. Hop...
- Wed Jan 22, 2020 1:39 am
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: 5J.13
- Replies: 5
- Views: 458
Re: 5J.13
You know that K decreases as temperature increases. This means that the reaction is exothermic. Because of this, the reactants are favored, and increasing the temperature will result in a shift to the reactant side. More ammonia will NOT be formed, because this is a product. You do not need an ICE t...
- Wed Jan 22, 2020 1:36 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: bar conversion
- Replies: 5
- Views: 205
Re: bar conversion
The conversion of bar to atm will be given in the formula handout we are given in each test. It is mainly a difference in values of R. If you are given values in bar, use the value for R involving bar, and then convert to atm using the formula sheet.
- Wed Jan 22, 2020 1:32 am
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Le Chatelier's in relation to stability
- Replies: 3
- Views: 114
Re: Le Chatelier's in relation to stability
No; Le Chatelier's principle only refers to stressors such as changing pressure or temperature or concentration.
- Wed Jan 22, 2020 1:30 am
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: 6D.15 part b
- Replies: 3
- Views: 255
Re: 6D.15 part b
This is an example of a hydrated complex. We haven't learned this in class, so I wouldn't worry about it. It is a topic that combines coordination complexes being hydrated in water (recall that it typically has 6 binding sites, which is why 6 H2O's can bind).
Hope this helped!
Hope this helped!
- Wed Jan 22, 2020 1:28 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: POST ASSESSMENT PART 1B NUMBER 27
- Replies: 1
- Views: 88
Re: POST ASSESSMENT PART 1B NUMBER 27
You can ignore the amount of C(s), because this does not affect the equilibrium (it is a solid). First convert the moles of gaseous H2O to concentration (moles / L). This is your initial. You can then complete the ICE table accordingly.
Hope this helped!
Hope this helped!
- Mon Jan 13, 2020 10:12 pm
- Forum: Ideal Gases
- Topic: Topics for Test 1
- Replies: 17
- Views: 651
Re: Topics for Test 1
Practice problems are definitely helpful, but make sure to go over your notes. Also, when you take notes in class, make sure you pay special attention to the things Lavelle mentions but doesn't put up on the slides. He tends to include some questions that require information on these minute details,...
- Mon Jan 13, 2020 10:05 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 6B.5d
- Replies: 2
- Views: 77
Re: 6B.5d
First find the number of moles of KOH in 2 mL of 0.175M solution. Then, divide by 0.500 L to find the concentration of KOH. The concentration of KOH equals the concentration of OH- in the solution because KOH is a strong base and will dissociate completely into K+ and OH- ions. So, you can find pOH ...
- Mon Jan 13, 2020 10:01 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: 5J.5
- Replies: 3
- Views: 132
Re: 5J.5
a) the reactants are favored
b) the reactants are favored (ignore the carbon because it is a solid, so there are fewer moles of gas on the reactant side)
c) the reactants are favored
d) neither are favored (equal number of moles of gas on both sides)
e) reactants are favored
b) the reactants are favored (ignore the carbon because it is a solid, so there are fewer moles of gas on the reactant side)
c) the reactants are favored
d) neither are favored (equal number of moles of gas on both sides)
e) reactants are favored
- Mon Jan 13, 2020 9:56 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: K vs Kc
- Replies: 2
- Views: 70
Re: K vs Kc
I was also confused by this, but I'm assuming that you should use the most specific option. For example, if the question is specifically referring to concentration values, it would be best to use Kc rather than K, because that way you know it's specifically for concentration.
- Mon Jan 13, 2020 9:51 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: HW 5I.13
- Replies: 1
- Views: 198
Re: HW 5I.13
Cl 2 (g) <---> 2Cl (g) K = 1.0 * 10 -5 I 0.001 0 C - x + 2x E 0.001 - x 2x The Cl 2 conc. can be found by doing (2.0 * 10 -3 mol Cl 2 ) / (2.0 L) = 0.001 M Assuming you got this ICE table, you shoukd get the equation: 1.2 * 10 -7 = (4x 2 ) / (0.001 - x) You should then get the quadratic equation fol...
- Thu Jan 09, 2020 11:09 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: HW 5J.5
- Replies: 8
- Views: 160
Re: HW 5J.5
You have to look at the moles of reactants and products that are in GAS form. There is only one mole of gas on the reactant side because C is a solid. On the other hand, there are two moles of gas on the product side. So, the reactant will proceed to the reactants.
Hope this helped!
Hope this helped!
- Thu Jan 09, 2020 11:06 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 5.35
- Replies: 4
- Views: 204
Re: 5.35
Do we have to use bar as the unit to calculate equilibrium constant? You can use either bar or atm. The units of pressure only matter when you are converting to concentration and you need to use the PV = nRT equation. In that case, you'll have to use the correct R, which has different values based ...
- Thu Jan 09, 2020 11:01 pm
- Forum: Ideal Gases
- Topic: Solving for K when only given balanced equation [ENDORSED]
- Replies: 6
- Views: 234
Re: Solving for K when only given balanced equation [ENDORSED]
P has nothing to do with whether we have the values or not, it just indicates that we are talking about partial pressure instead of concentration. Usually, you use partial pressure when the substances in the reaction are gases (as opposed to aqueous solutions).
Hope this helped!
Hope this helped!
- Thu Jan 09, 2020 10:59 pm
- Forum: Ideal Gases
- Topic: Adding reactions
- Replies: 4
- Views: 239
Re: Adding reactions
It's a similar concept as determining the formula for K. For example, if you have Br2Cl2 --> Br2 + Cl2, you would write K as: K = [Br2][Cl2] / [Br2Cl2] Even though you're adding Br2 and Cl2 in the reaction, you multiply them in K. So, adding translates to multiplying when calculating K. Hope this he...
- Thu Jan 09, 2020 10:57 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Difference between -> and <->
- Replies: 4
- Views: 169
Re: Difference between -> and <->
I assume you're referring to that problem on Q in the textbook. Generally, for Q, you are observing a reaction at a certain point, so you are only looking at a certain direction of the reaction
Hope this helped!
Hope this helped!
- Wed Dec 04, 2019 11:21 am
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: What is s-character?
- Replies: 1
- Views: 181
Re: What is s-character?
s-character gives the contribution of the s-orbital to the bond (sp 3 = 25%, sp 2 = 33%, sp = 50%, etc.). As s-character increases, bond angle also increases (lone pairs decrease bond angle and decrease s-character because there are more regions of electron density and the s-orbital does not contrib...
- Wed Dec 04, 2019 11:17 am
- Forum: Properties & Structures of Inorganic & Organic Acids
- Topic: D-block metal, character of oxides
- Replies: 1
- Views: 93
Re: D-block metal, character of oxides
Higher oxidation states of transition metals result in a higher electron density. The d-block metal pulls the electrons in the bond with the oxygen toward itself, making the bond between the oxygen and the hydrogen weaker. The hydrogen can thus be removed more easily, making it a stronger acid. Hope...
- Wed Dec 04, 2019 11:14 am
- Forum: Hybridization
- Topic: Unhybridized Orbitals and pi bonds
- Replies: 2
- Views: 124
Re: Unhybridized Orbitals and pi bonds
Yes. A pi bond is specifically made up of unhybridized p-orbitals that overlap side by side. Sigma bonds are formed by hybridized orbitals.
Hope this helped!
Hope this helped!
- Wed Dec 04, 2019 11:13 am
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: ICE tables
- Replies: 3
- Views: 252
Re: ICE tables
If x is less than 5% of the initial value, you can ignore the " - x" step in the ICE table because the subtraction is negligible. If it is over 5%, you must account for this subtraction.
Hope this helped!
Hope this helped!
- Wed Dec 04, 2019 11:12 am
- Forum: Naming
- Topic: Determining Type of Ligand
- Replies: 2
- Views: 76
Re: Determining Type of Ligand
Only the number of atoms with the lone pair count toward the coordination number. For example, water is monodentate, even though oxygen has 2 lone pairs. The transition metal will bond to the one oxygen atom, not the individual lone pairs.
Hope this helps!
Hope this helps!
- Mon Nov 25, 2019 11:36 pm
- Forum: Bronsted Acids & Bases
- Topic: J.1
- Replies: 3
- Views: 235
Re: J.1
For starters, you should memorize the strong acids (HBr, Hi, HCl, HClO3, HClO4, H2SO4, HNO3) and the strong bases (LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, Ba(OH)2. This will help you figure out a lot of the Bronsted acids and bases. It also helps to know conjugate acid/base pairs, such as NH4...
- Mon Nov 25, 2019 11:32 pm
- Forum: Ionic & Covalent Bonds
- Topic: Intra vs Inter molecular forces
- Replies: 7
- Views: 1353
Re: Intra vs Inter molecular forces
Intermolecular forces are between molecules and they are the forces of attractions between separate molecules that cause them to form liquids and solids. Intramolecular forces are forces which hold atoms together to form a molecule (like a chemical bond). As Michelle said, INTERmolecular = between ...
- Mon Nov 25, 2019 11:29 pm
- Forum: Properties & Structures of Inorganic & Organic Bases
- Topic: strength of base
- Replies: 8
- Views: 825
Re: strength of base
For halogen acids (H-X, where X is a halogen), the atomic radius of X will result in a stronger bond between H and X. This will weaken the acid because it won't dissociate as easily in water. As a result, it will be a weaker acid and a stronger base. For oxyacids (H-O-X, where X is a group 5A, 6A, o...
- Mon Nov 25, 2019 11:23 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Online Textbook
- Replies: 6
- Views: 354
Re: Online Textbook
Some people have the PDF for the seventh edition Chemistry textbook. Maybe someone can link it?
- Mon Nov 25, 2019 11:22 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Bond type and coordinate compounds
- Replies: 1
- Views: 211
Re: Bond type and coordinate compounds
From what I remember from lecture, monodentate ligands are just ligands that do not have more than one bonding site. This has nothing to do with forming single or double bonds.
In other words, there is nothing stopping a pi bond from being polydentate.
Hope this helped!
In other words, there is nothing stopping a pi bond from being polydentate.
Hope this helped!
- Mon Nov 18, 2019 2:44 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: CH2Cl2 structure
- Replies: 2
- Views: 170
Re: CH2Cl2 structure
Either can occur; they are both equally likely structures. I think it'll just depend on the question. For example, if the question asks you to draw the Lewis structure such that C 2 H 2 Cl 2 is polar, you would draw the structure so that the Cl atoms are bonded to the carbons on one side and the H a...
- Mon Nov 18, 2019 2:41 pm
- Forum: Hybridization
- Topic: hybridization 1st number
- Replies: 3
- Views: 251
Re: hybridization 1st number
The number refers to the outermost energy level. In the example you showed, the outermost energy level, where all the valence electrons are residing, is n = 2. So, the valence electrons from this ring will interact and form hybrid orbitals.
Hope this helped!
Hope this helped!
- Mon Nov 18, 2019 2:39 pm
- Forum: Electronegativity
- Topic: test
- Replies: 7
- Views: 518
Re: test
As you go from left to right, electronegativity will increase (aka it will take require more energy to remove an electron). This is because the valence electrons are experiencing a stronger pull from the nucleus. As you go from top to bottom, electronegativity decreases because the valence electrons...
- Mon Nov 18, 2019 2:34 pm
- Forum: Hybridization
- Topic: Where does hybridization occur?
- Replies: 1
- Views: 73
Re: Where does hybridization occur?
For our purposes, all single bonds are hybridized. These are called sigma bonds. This is not a phenomenon; rather, it is an accepted part of chemical structures that in order for covalent molecules to share electrons, they must have hybrid orbitals. For double bonds, one bond will be hybridized, and...
- Mon Nov 18, 2019 2:29 pm
- Forum: Hybridization
- Topic: Relationship between VSEPR, hybridization, and Lewis
- Replies: 1
- Views: 72
Re: Relationship between VSEPR, hybridization, and Lewis
That's correct. By drawing the Lewis structure, we can determine the VSEPR model (shape/ angle of bond/ general VSEPR formula). We can then determine the hybridization of the central atom by looking at the number of bonding pairs and lone pairs surrounding the central atom. This is why we built up t...
- Mon Nov 18, 2019 2:26 pm
- Forum: Dipole Moments
- Topic: 3F.1 Dipole-dipole vs dipole-induced dipole
- Replies: 1
- Views: 94
Re: 3F.1 Dipole-dipole vs dipole-induced dipole
H 2 SeO 4 has hydrogen bonding, because it has an OH bond. To differentiate between dipole-dipole and dipole-induced dipole, look at polarity. If the molecule is polar, it will do dipole-dipole bonding because the differences in electronegativity will allow attractions between opposite forces. If it...
- Tue Nov 12, 2019 11:42 am
- Forum: Bond Lengths & Energies
- Topic: Hw Problem 3F.19
- Replies: 1
- Views: 184
Re: Hw Problem 3F.19
First determine the types of intermolecular forces solid Xe and solid Ar have. Neither are polar (considering it's just a single atom), so they cannot do dipole-dipole or hydrogen bonding. This leaves us with London dispersion forces. When determining the strength of London dispersion forces, look a...
- Tue Nov 12, 2019 11:38 am
- Forum: Dipole Moments
- Topic: Melting Point Due to Anion Size
- Replies: 1
- Views: 125
Re: Melting Point Due to Anion Size
Ionic and covalent bonds are relevant as INTRAmolecular attractions (within the same molecule). However, melting point concerns INTERmolecular attractions (between two molecules). When something melts, its individual molecules remain the same, but the bond between different molecules is weakened. If...
- Tue Nov 12, 2019 11:34 am
- Forum: Bond Lengths & Energies
- Topic: Vapor Pressure
- Replies: 1
- Views: 157
Re: Vapor Pressure
As boiling point increases, vapor pressure increases. If intermolecular attractions in the liquid are weak, its vapor pressure is high (it will vaporize easily). However, if the intermolecular attractions are strong, it will require more heat to break them, and so the boiling point will be higher. H...
- Tue Nov 12, 2019 10:25 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: VSEPR
- Replies: 5
- Views: 110
Re: VSEPR
We will probably eventually have to know the names, as well as the expected bond angles, of VSEPR. Further, we'll have to know the hybridization, or hybrid orbitals, of a given atom in a molecule. But Dr. Lavelle will probably go over all this in a lot of depth before expecting us to know it. Hope t...
- Tue Nov 12, 2019 10:23 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Homework for Week 7
- Replies: 5
- Views: 288
Re: Homework for Week 7
I think we should mainly focus the homework on 3F, because this talks about intermolecular attractions, and we just started this topic in lecture. As we get into VSEPR, we'll be able to use problems from the next unit.
Hope this helped!
Hope this helped!
- Tue Nov 12, 2019 10:22 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Forces
- Replies: 3
- Views: 92
Re: Forces
I haven't heard that, but if I had to guess, I would say it's because of the electron-electron repulsion. The atom with the lone pairs will push away from the atom it's connected to because the lone pairs exert a repulsive force. I have heard that the bond angle will be smaller for this reason. For ...
- Fri Nov 08, 2019 1:09 am
- Forum: Administrative Questions and Class Announcements
- Topic: Midterm --> Final Concepts
- Replies: 3
- Views: 225
Midterm --> Final Concepts
Does anyone know if we'll be tested on the midterm concepts in the final? Is the final a cumulative of all concepts we've learned throughout the course? If so, I'd like to keep practicing the ideas we covered on the midterm so I don't forget them by the final :/
Thank you!
Thank you!
- Fri Nov 08, 2019 1:05 am
- Forum: Dipole Moments
- Topic: Discussion Worksheet
- Replies: 1
- Views: 104
Re: Discussion Worksheet
Anions that are very polarizable will be large and electron-rich. This means they will have a low charge. The order is O2- < N3- < Cl- < Br- because O2- is electron-rich and the smallest anion given. Br- is the largest anion given (significantly larger than Cl- because of its extra shell), and is th...
- Fri Nov 08, 2019 1:03 am
- Forum: Lewis Structures
- Topic: Midterm grades
- Replies: 26
- Views: 1370
Re: Midterm grades
Dr. Lavelle said because there are so many students in the class, it should take about a week (maybe a day or two longer). I'm pretty sure the TA's are grading our papers, so it shouldn't take over two weeks.
- Thu Nov 07, 2019 2:32 pm
- Forum: Administrative Questions and Class Announcements
- Topic: What homework to turn in per week
- Replies: 7
- Views: 361
Re: What homework to turn in per week
It depends on how long we spend on a unit. If the concept is still relevant in lecture, you can do questions on it and turn it in for credit.
Hope this helped!
Hope this helped!
- Thu Nov 07, 2019 2:32 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: VESPR
- Replies: 5
- Views: 344
Re: VESPR
A lot of high school chemistry classes taught VSEPR; however, don't worry if you don't know them. I'm sure Dr. Lavelle is going to go over it with us. The table is in the textbook I think.
Hope this helped!
Hope this helped!
- Sun Nov 03, 2019 9:46 pm
- Forum: Dipole Moments
- Topic: Why can water break up otherwise strong Ionic bonds?
- Replies: 1
- Views: 302
Re: Why can water break up otherwise strong Ionic bonds?
The hydrogen bond is very strong (stronger than a dipole-dipole interaction) because of the huge difference in electronegativity. When ionic compounds are dissolved in water, the positive dipole end of the water molecule (H+) attracts the anion from the compound, while the negative dipole end of ano...
- Sun Nov 03, 2019 9:43 pm
- Forum: Octet Exceptions
- Topic: Uhhh
- Replies: 3
- Views: 140
Re: Uhhh
P, S, and Cl can accommodate more than 8 valence electrons. This is because atoms in period 3 or higher have d-orbitals in the valence shell that can accommodate additional electrons. These are all the octet exceptions, to my knowledge.
Hope this helped!
Hope this helped!
- Sun Nov 03, 2019 9:41 pm
- Forum: Lewis Structures
- Topic: Showing work for valence electrons
- Replies: 3
- Views: 164
Showing work for valence electrons
This is more of a concern for tests – do we have to show work when calculating valence electrons? Will Dr. Lavelle care if we just write down the number of valence electrons when trying to determine the Lewis structure of a molecule?
Thanks!
Thanks!
- Tue Oct 29, 2019 5:55 pm
- Forum: Dipole Moments
- Topic: 3F.5
- Replies: 1
- Views: 101
Re: 3F.5
There are no hydrogen bonds in CHF3. For there to be a hydrogen bond, the F and H must actually be connected to each other. However, the Lewis structure of CHF3 shows C as the central atom, with the three F's and the one H all connected to the C. So, we only look at London dispersion forces in the t...
- Tue Oct 29, 2019 5:50 pm
- Forum: Dipole Moments
- Topic: intermolecular interactions
- Replies: 2
- Views: 163
Re: intermolecular interactions
A molecule can do dipole-dipole bonding if it is polar. It can do hydrogen bonding if it is polar and if it has an OH- or NH- bond. H-bonding is basically a type of dipole-dipole bonding, except with OH and NH bonds. All molecules have London dispersion forces, which are momentary attractive charges...
- Tue Oct 29, 2019 5:43 pm
- Forum: *Particle in a Box
- Topic: En =((h^2)(n^2))/(8 m L^2)
- Replies: 1
- Views: 390
Re: En =((h^2)(n^2))/(8 m L^2)
No, we have not covered this equation in class. If we do go over it, I expect it will be later in the course. However, considering it's related to energy, we might just not cover it, as we're already finished with the quantum unit.
- Tue Oct 29, 2019 5:37 pm
- Forum: Lewis Structures
- Topic: Chemistry Terminology for Electron and Lone Pairs
- Replies: 1
- Views: 86
Re: Chemistry Terminology for Electron and Lone Pairs
It depends on what orbital the electrons are in. For example, carbon has 4 valence electrons, but they are all in their own orbitals, so none of them are paired. However, if two electrons are in the same orbital and not bonded to another atom, then these are lone pairs. Bonding pairs would be the &q...
- Mon Oct 28, 2019 1:40 pm
- Forum: Lewis Structures
- Topic: Expanded Valence Shells
- Replies: 3
- Views: 122
Expanded Valence Shells
I remember in AP Chem, we learned that P, S, Cl, Br, I, and Xe all had expanded valence shells. This is because the outer shells are very large (atomic radius is large), and can accommodate more than 8 electrons. However, Dr. Lavelle only mentioned P, S, and Cl as having expanded valence shells. Is ...
- Mon Oct 28, 2019 1:37 pm
- Forum: Ionic & Covalent Bonds
- Topic: Ground-State electron configuration
- Replies: 3
- Views: 157
Re: Ground-State electron configuration
When an electron is in its ground-state, this means it is in its lowest-possible energy level. When the electron is excited, it absorbs energy and jumps to an energy level higher than the ground state energy level. Usually, the electron is not stable when it is in an excited state, and will eventual...
- Wed Oct 23, 2019 1:23 pm
- Forum: Resonance Structures
- Topic: polarity
- Replies: 1
- Views: 98
Re: polarity
Resonance structures have delocalized electrons, which means the negative charge is spread throughout all the bonds in the molecule. This affects polarity, as the spread charges reduce the polarity of the overall molecule.
Hope this helped!
Hope this helped!
- Wed Oct 23, 2019 1:15 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: 1E.22
- Replies: 1
- Views: 104
Re: 1E.22
a) Pb --> The valence shell is 6p, and there are 2 unpaired electrons in this shell. b) Ir --> there are 2 unpaired electrons (5 orbitals in valence shell, 7 electrons in the shell) c) Y --> 1 unpaired electron in valence shell (4d) d) Cd --> no unpaired electrons (5 orbitals in valence shell 4d, 10...
- Tue Oct 22, 2019 5:32 pm
- Forum: Properties of Electrons
- Topic: x,y,z for Electrons
- Replies: 10
- Views: 381
Re: x,y,z for Electrons
Adding on to all the previous comments, the X, Y, and Z in the electron configuration indicate the different planes. I also think Dr. Lavelle said it was a good idea to specify this because it shows that the electrons are unpaired. For example, when you write 1s2 2s2 3px1 3py1 3pz1, you are showing ...
- Tue Oct 22, 2019 5:26 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Electron Configuration of Tantalum
- Replies: 1
- Views: 220
Re: Electron Configuration of Tantalum
For tantalum, the entire 4f orbital is completely filled (14 electrons). We know this because the f-block falls in the periodic table before we reach tantalum. The 6s orbital is completely filled as well; this just follows the same pattern we've been using for d-block elements (besides the exception...
- Tue Oct 22, 2019 3:21 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Knowing the nm of light
- Replies: 2
- Views: 86
Re: Knowing the nm of light
I remember my TA saying something about knowing the general range for visible light and UV radiation. Even this doesn't need to be too specific. 400-700nm is the visible spectrum, which might be a good thing to memorize so you can gauge the general range for the other wavelengths. Overall, I wouldn'...
- Wed Oct 16, 2019 1:17 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Quantum Numbers
- Replies: 7
- Views: 233
Re: Quantum Numbers
The quantum numbers all describe a single electron in any atom. So it doesn't have to be an atom with ONLY one electron; Lavelle was just saying that the four numbers all describe a single electron – the energy, shape, orientation, and spin state.
Hope this helped!
Hope this helped!
- Wed Oct 16, 2019 1:12 pm
- Forum: Accuracy, Precision, Mole, Other Definitions
- Topic: Question regarding definition of molecules
- Replies: 5
- Views: 530
Re: Question regarding definition of molecules
It can also be used to find photons per mole! This conversion is probably good to know, especially for the quantum mechanics unit :)
1 mole = 6.02 * 10^23 photons
1 mole = 6.02 * 10^23 photons
- Wed Oct 16, 2019 1:10 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Electron spin
- Replies: 8
- Views: 274
Re: Electron spin
I don't think we will be going into finding the spin state of a specific electron. The idea of that quantum number is just that an electron can either have a positive spin or a negative spin, or a quantum number of +1/2 or -1/2. I think the main concept to understand is that if given 2 electrons in ...
- Wed Oct 16, 2019 1:07 pm
- Forum: SI Units, Unit Conversions
- Topic: Use of Angstrom?
- Replies: 3
- Views: 293
Re: Use of Angstrom?
Based on AP Chemistry, we rarely had to convert to units of Angstrom. That being said, Angstroms are most used to describe bond length, which I believe is our next unit. We may be expected to use it most in that unit.
Hope this helped!
Hope this helped!
- Wed Oct 16, 2019 1:05 pm
- Forum: Properties of Light
- Topic: Difference between Quanta and photons?
- Replies: 6
- Views: 617
Re: Difference between Quanta and photons?
I think a "quantum" is an amount, whereas a photon is an actual bundle of energy. This is why a photon is also known as a "quantum of energy".
Hope this helped!
Hope this helped!