Are we expected to know how to come up with the half reactions for reactions such as those in 14.13 part D and 14.15 part A (6th edition)?
Or can we only get the half reaction if the appendix is given?
Search found 72 matches
- Wed Mar 13, 2019 7:46 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Cell Diagram Half Reactions
- Replies: 1
- Views: 278
- Mon Mar 11, 2019 10:26 pm
- Forum: Second Order Reactions
- Topic: Slope of k
- Replies: 2
- Views: 367
Slope of k
Why is the slope, k, of second order reactions positive?
- Mon Mar 11, 2019 4:17 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Homework 15.63
- Replies: 2
- Views: 396
Re: Homework 15.63
You use the Arrhenius equation for this problem. ln(k2/k1) = Ea/R(1/T1-1/T2)
Given:
Ea = 38 kj/mol
k1 = 1.5E10 L/mols
T1 = 298 K
k2 = ?
T2 = 310 K
lnk2 = ln(1.5E10) + (38/.008314)(1/298 - 1/310)
k2 = e^24 = 2.7E10 L/mols
Given:
Ea = 38 kj/mol
k1 = 1.5E10 L/mols
T1 = 298 K
k2 = ?
T2 = 310 K
lnk2 = ln(1.5E10) + (38/.008314)(1/298 - 1/310)
k2 = e^24 = 2.7E10 L/mols
- Mon Mar 11, 2019 4:09 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Question 15.5 (Sixth Edition)
- Replies: 2
- Views: 812
Re: Question 15.5 (Sixth Edition)
Remember that the unique rate for aA + bB --> cC + dD is -1/a(d[A]/d[t]) + -1/b(d[B]/d[t]) --> 1/c(d[C]/d[t]) + 1/d(d[D]/d[t]). You are given the unique rate 0.44 mol/Ls. In order to find (a), you set 0.44 = 1/3(d[A]/d[t]), so to solve for (d[A]/d[t]), 0.44 * 3 = 1.3 mol O2/Ls. You do the same for w...
- Mon Mar 11, 2019 4:02 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Homework 15.65
- Replies: 1
- Views: 353
Re: Homework 15.65
(a) In order to find the equilibrium constant, you divide the forward rate constant by the reverse rate constant. K = k1/k'1 = 265/392 = 0.676. (b) The reaction is endothermic because the activation energy of the forward reaction is higher than the activation energy of the reverse reaction. In an en...
- Sun Mar 10, 2019 10:48 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Bond Enthalpies in the Gas Phase
- Replies: 1
- Views: 304
Bond Enthalpies in the Gas Phase
When calculating enthalpy using bond enthalpies, do the calculations apply only for gaseous substances? If so, how do you know when to add or subtract the enthalpy of vaporization to change a liquid reactant or product?
- Mon Mar 04, 2019 10:38 am
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Integrals/derivatives
- Replies: 3
- Views: 471
Re: Integrals/derivatives
I believe knowing how to solve integrals and derivatives are mainly for us to know how to derive certain equations in this class. So far, we have been using integrals and derivatives to get formulas such as that of work of an reversible, isothermal reaction, change in entropy, work of a system with ...
- Mon Mar 04, 2019 10:32 am
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: 15.3
- Replies: 1
- Views: 337
Re: 15.3
You divide what you got in part (a) by 2 because there are 2 mols of NO2 in the reaction. Another way is you can also divide (b) by one because there is one mol of O2 formed. You divide the rate of reaction by the stoichiometric coefficient to get 3.3e-3 mol/Ls. If you have aA --> bB + cC, the uniqu...
- Mon Mar 04, 2019 10:25 am
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: 14.101 6th edition
- Replies: 1
- Views: 302
Re: 14.101 6th edition
For this problem you will use the equation E = Ecell - (RT/nF)ln[K]. In this problem, you would look at it as E = Ecell - (RT/nF)ln[(Kout)/(Kin)] Because both half cells are identical, you should expect Ecell = 0. In order to find the range, you have to first calculate how E would be when ions are 3...
- Mon Feb 25, 2019 1:48 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Anode and Cathode
- Replies: 6
- Views: 627
Re: Anode and Cathode
Whichever ion undergoes oxidation is the anode and whichever ion undergoes reduction is the cathode.
- Mon Feb 25, 2019 11:10 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Gibbs Free Energy
- Replies: 1
- Views: 262
Re: Gibbs Free Energy
A compound is stable if the delta G of the reaction is < 0. This means that the reaction is spontaneous and proceeds to the forward reaction so the products are stable. A reaction is unstable if delta G > 0. Stability and unstability are related to the tendency of a reaction to decompose into its el...
- Mon Feb 25, 2019 10:22 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Question 14.19 (Sixth Edition)
- Replies: 1
- Views: 422
Re: Question 14.19 (Sixth Edition)
We know that Cu/Cu2+ is the anode. The standard potential for Cu/Cu2+ which can be found in the Appendix, is +0.34.
You plug this into the E(cell) = E(cathode) - E(anode) equation to find the potential for the cathode, M2+/M.
E(cathode) = E(cell) + E(anode) = -0.689 + 0.34 = -0.349 V.
You plug this into the E(cell) = E(cathode) - E(anode) equation to find the potential for the cathode, M2+/M.
E(cathode) = E(cell) + E(anode) = -0.689 + 0.34 = -0.349 V.
- Sat Feb 23, 2019 3:04 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Standard Potentials and Equilibrium Constants
- Replies: 2
- Views: 309
Standard Potentials and Equilibrium Constants
Will we need to know for the test how to derive the equilibrium constant for a reaction given electrochemical data? (Section 14.8 in the textbook)
- Tue Feb 19, 2019 7:06 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 4I.9
- Replies: 2
- Views: 543
Re: 4I.9
Remember that in a reversible reaction, any small change in a variable, such as pressure, of the system will cause a similar change in the surroundings so that there is essentially no change in internal energy. ( U = 0 ) Because in part b the pathway is irreversible, this means that any change in va...
- Tue Feb 19, 2019 12:05 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Concentration and free energy
- Replies: 2
- Views: 432
Concentration and free energy
Does changing the concentration of a substance in a reaction affect the free energy?
- Tue Feb 19, 2019 11:03 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Relationship between enthalpy, entropy, and Gibbs
- Replies: 4
- Views: 538
Relationship between enthalpy, entropy, and Gibbs
How does free energy change with temperature when the value of enthalpy is negative and entropy is negative?
How does free energy change with temperature when the value of enthalpy is positive and entropy is positive?
How does free energy change with temperature when the value of enthalpy is positive and entropy is positive?
- Mon Feb 11, 2019 10:23 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Question 8.39 (Sixth Edition)
- Replies: 4
- Views: 514
Re: Question 8.39 (Sixth Edition)
For this problem, you have to calculate the enthalpy of fusion (ice melting) and the heat of the water to determine how much needs to be supplied in total. The enthalpy of fusion, delta H, can be calculated using the value from table 8.3. Convert 80.0 g H20 to moles and multiply it with the enthalpy...
- Mon Feb 11, 2019 10:09 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Question 8.19 (Sixth Edition)
- Replies: 4
- Views: 1525
Re: Question 8.19 (Sixth Edition)
So with this question, you want to find q of both the copper and the water and add it together to find the total amount of heat supplied. You know that q = mCdeltaT q(copper) = (500 g)(0.38 J/gC)(100-22 C) = 14.82 kJ q(water) = (400 g)(4.184 J/gC)(100-22 C) = 130.54 kJ q(total) = 14.82 + 130.54 = 14...
- Mon Feb 11, 2019 10:03 am
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: 12.55 part b 6th edition
- Replies: 1
- Views: 484
Re: 12.55 part b 6th edition
You should plug it in as -0.3+√(0.3^2 - 4(1)(-0.6)) and divide it all by 2. You should get a positive number inside the square root and x = 0.137.
- Tue Feb 05, 2019 4:44 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Equipartition Theorem
- Replies: 1
- Views: 300
Equipartition Theorem
Are we required to know the equipartition theorem?
I'm also confused about the translational, rotational linear, and rotational nonlinear equations that go along with it.
I'm also confused about the translational, rotational linear, and rotational nonlinear equations that go along with it.
- Mon Feb 04, 2019 10:26 am
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: 8.41 6th edition
- Replies: 2
- Views: 316
Re: 8.41 6th edition
In this problem, you add the enthalpy of fusion, heat absorbed by ice, and heat released by water. In order to find the enthalpy of fusion, you convert 50.0 g H20(s) to mol and multiply it by the enthalpy of fusion. This would be 2.775 mol H20 * 6.01E3 J/mol (convert kJ in textbook to J). The heat a...
- Mon Feb 04, 2019 10:15 am
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Open vs Closed Systems
- Replies: 4
- Views: 512
Re: Open vs Closed Systems
An open system is one in which matter and energy can exchange with the surroundings. A closed system is one in which only energy can exchange with the surroundings. An isolated system is one in which nothing can exchange with the surroundings. For example, an open system can just be a cup of hot wat...
- Mon Feb 04, 2019 10:04 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Bond Enthalpy
- Replies: 2
- Views: 327
Re: Bond Enthalpy
I'm assuming mostly like no. We learned lewis structures in 14A so I think they expect you to be able to know which bonds would be broken or formed.
- Tue Jan 29, 2019 10:26 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Kekulé Structures
- Replies: 1
- Views: 286
Re: Kekulé Structures
For this problem, they are just asking you to compare the differences between the bond enthalpies of benzene. In one structure, you have benzene with 3 C-C and 3 C=C bonds. Using the table, this would mean (3*348) + (3*612) = 2880 kJ. However, benzene in its resonance structure has 6 C-C/C=C bonds (...
- Tue Jan 29, 2019 11:26 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Sig Figs with Enthalpy
- Replies: 4
- Views: 661
Sig Figs with Enthalpy
Do we apply the rules of sig figs when calculating enthalpy?
- Tue Jan 29, 2019 11:24 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Heat and work
- Replies: 4
- Views: 468
Heat and work
What is the difference between heat and work? How do they relate to internal energy?
- Tue Jan 22, 2019 12:02 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 5G.3 7th Edition (When to use P vs molecular formula/concentrations)
- Replies: 5
- Views: 477
Re: 5G.3 7th Edition (When to use P vs molecular formula/concentrations)
You can use either one.
- Tue Jan 22, 2019 11:18 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 5G.3 7th Edition (When to use P vs molecular formula/concentrations)
- Replies: 5
- Views: 477
Re: 5G.3 7th Edition (When to use P vs molecular formula/concentrations)
The solution is just writing out the chemical equilibrium constant in the form of Kp rather than Kc. They are interchangeable for gases only. When writing it with P subscripts, this indicates that it is Kp, with the equilibrium partial pressures of products on top divided by the equilibrium partial ...
- Tue Jan 22, 2019 10:22 am
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Temperature change
- Replies: 5
- Views: 592
Re: Temperature change
The increase in temperature depends on whether the reaction is exothermic or endothermic. If a reaction is endothermic, an increase in temperature will make it proceed to the right and produce more products. This means the reverse reaction is exothermic, so a decrease in temperature will make it pro...
- Tue Jan 22, 2019 10:16 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Test and acid/ base recognition
- Replies: 1
- Views: 166
Re: Test and acid/ base recognition
From Chem 14A, we had to memorize which acids were strong and what categorized bases as strong. Strong acids: HCl, HI, HBr, H2SO4, HNO3, HClO4 Strong bases: Most of Group 1 and 2 metal hydroxides If you know those, then everything else should be a weak acid or base. However, I would assume in the qu...
- Thu Jan 17, 2019 10:47 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Hemoglobin and Homeostasis
- Replies: 1
- Views: 142
Hemoglobin and Homeostasis
How does partial pressure and concentration of oxygen affect hemoglobin and oxygen exchange?
- Tue Jan 15, 2019 2:00 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Part 4 Post-Module Question
- Replies: 4
- Views: 1285
Re: Part 4 Post-Module Question
In this problem, we are taking into account of exothermic and endothermic reactions. The reaction shifts toward the endothermic reaction. If a reaction is endothermic, it will have a positive delta H, meaning that increasing temperature will favor the forward reaction and product formation. If a rea...
- Tue Jan 15, 2019 1:54 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: positive vs. negative delta H
- Replies: 2
- Views: 44359
Re: positive vs. negative delta H
Delta H basically means the change in enthalpy, or in other words, the change in "heat". When delta H is positive, it means the products in the reaction have more energy compared to the reactants, so the reaction has gained energy, making it endothermic. When delta H is negative, it means ...
- Tue Jan 15, 2019 1:50 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Hmwrk 11.41 6th edition
- Replies: 1
- Views: 380
Re: Hmwrk 11.41 6th edition
The answer is 1.58E-8. In order to do this problem you convert 17.4 mg CO2 to moles and divide by 0.250 L to get the equilibrium molarity concentration of CO2, 1.58E-3. Since NH4(NH2CO2) is a solid, it's not included in the equilibrium constant, so you have K = [NH3]^2[CO2]. Because you have two mol...
- Thu Jan 10, 2019 11:09 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Help with question 11.43 6th edition [ENDORSED]
- Replies: 2
- Views: 251
Re: Help with question 11.43 6th edition [ENDORSED]
On the ICE table, you have to find initial molar concentration, change in molar concentration, and the equilibrium molar concentration. Since you are initially given 1.0 mol/L NO, the change would be -2x since you have the coefficient 2NO in the reaction. In order to calculate equilibrium, you subtr...
- Wed Jan 09, 2019 9:47 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 6th Edition 11.45
- Replies: 1
- Views: 165
6th Edition 11.45
Hello! I'm stuck on question 11.45 part b. (a) A sample of 2.0 mmol Cl2 was sealed into a 2.0 L reaction vessel and heated to 1000. K to study its dissociation into Cl atoms. Use the info in Table 11.2 to calculate the equilibrium composition of the mixture. (b) If 2.0 mmol F2 was placed into the re...
- Wed Jan 09, 2019 5:46 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 11.9 Hmwrk 6th edition
- Replies: 1
- Views: 387
Re: 11.9 Hmwrk 6th edition
Yes; since you aren't given any numerical equilibrium molar concentrations, the question just asks for you to set it up.
So basically it would be Kc = [products]/[reactants] raised to the power of their stoichiometric coefficients.
So basically it would be Kc = [products]/[reactants] raised to the power of their stoichiometric coefficients.
- Mon Dec 03, 2018 4:39 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: 6th Edition 17.37
- Replies: 2
- Views: 355
Re: 6th Edition 17.37
For (c), en is a bidentate, so it has two binding sites. (en)2 = 4 binding sites total + 2 binding Cl = 6 ligands total, so the coordination number is 6. For (d), edta is a hexadentate, so it has six binding sites, making the coordination number for Cr to be 6. More info can be found in section 17.6...
- Mon Dec 03, 2018 3:47 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Ch.17-#31b (6th edition) Coordination Compounds
- Replies: 2
- Views: 336
Re: Ch.17-#31b (6th edition) Coordination Compounds
Yes, ligands should be written in alphabetical order, ignoring the prefix that indicates the number of each one present.
For example, in that particular compound, "ammine" comes before "sulfato" in alphabetical order, so NH3 is written first.
For example, in that particular compound, "ammine" comes before "sulfato" in alphabetical order, so NH3 is written first.
- Mon Dec 03, 2018 3:12 pm
- Forum: Naming
- Topic: Prefix bis, tris, tetrakis, etc. [ENDORSED]
- Replies: 5
- Views: 772
Prefix bis, tris, tetrakis, etc. [ENDORSED]
Hello!
I'm a little confused as to when we name coordination compounds with bis, tris, tetrakis, etc.
What's an example of when this is used instead of di, tri, tetra, etc.?
Thank you!
I'm a little confused as to when we name coordination compounds with bis, tris, tetrakis, etc.
What's an example of when this is used instead of di, tri, tetra, etc.?
Thank you!
- Sun Dec 02, 2018 1:14 pm
- Forum: Lewis Acids & Bases
- Topic: 6th Edition 12.15
- Replies: 1
- Views: 237
6th Edition 12.15
Hello! Question 12.15 in the 6th edition asks to write the Lewis structure of each reactant, identify the Lewis acid and the Lewis base, and then write the Lewis structure of the product for the following acid-base reaction: (b) Cl- + SO2 --> I'm a little confused on this one, because I thought SO2 ...
- Mon Nov 26, 2018 10:00 am
- Forum: Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)
- Topic: Water as a solvent
- Replies: 2
- Views: 526
Re: Water as a solvent
Water has a bent shape and this makes it a universal solvent because the positive and negative ends are furthest from each other. Since the H atoms give off a positive charge and oxygen gives off a negative charge, molecules of either charge are able to attract to water. Water's polar covalent chara...
- Mon Nov 26, 2018 9:48 am
- Forum: Hybridization
- Topic: Hybridization with Lone Pairs
- Replies: 4
- Views: 488
Re: Hybridization with Lone Pairs
Yes, in this case N would be sp3 hybridized. When a lone pair of electrons are localized, they are considered in hybrid orbitals.
However, in the case of resonance structures, delocalized electrons in the resonance structures are not counted in hybrid orbitals.
However, in the case of resonance structures, delocalized electrons in the resonance structures are not counted in hybrid orbitals.
- Mon Nov 26, 2018 9:45 am
- Forum: Bond Lengths & Energies
- Topic: Covalent Bond Dissociation Energy
- Replies: 2
- Views: 314
Covalent Bond Dissociation Energy
What exactly is covalent bond dissociation energy and how does it relate to atomic radius, bond multiplicity, and lone pairs in bonding?
- Tue Nov 20, 2018 2:10 pm
- Forum: Electronegativity
- Topic: Electronegativity and Polarity
- Replies: 3
- Views: 462
Electronegativity and Polarity
What's the relation between polarizability and electronegativity?
- Tue Nov 20, 2018 1:48 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: lone pairs
- Replies: 4
- Views: 451
Re: lone pairs
I believe lone pairs on non central atoms would affect the polarity of the molecule and create a dipole, but it shouldn't affect the overall shape of the molecule. For example, with CH2F2 has a tetrahedral shape. However, F has lone pairs. This would just mean that the molecule is polar, but the sha...
- Tue Nov 20, 2018 1:38 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Hund's rule vs Pauli's
- Replies: 3
- Views: 1103
Re: Hund's rule vs Pauli's
Hund's rule states that one electron must fill each orbital first with a parallel spin before it can fill up with two electrons in each orbital with opposite spins. For example, in a 2p orbital, one electron must fill each 2px1, 2py1, 2pz1, before another electron can fill those orbitals. The Paul E...
- Tue Nov 13, 2018 10:07 pm
- Forum: Dipole Moments
- Topic: Induced-dipole
- Replies: 5
- Views: 1854
Re: Induced-dipole
Dipoles are unequal pulls of electrons. Induced-dipoles happen in London dispersion forces. What happens is that in an atom with an equal amount of electrons dispersed around, a temporary dipole force can occur. For instance, some electrons can temporarily shift to one side, causing one side of the ...
- Tue Nov 13, 2018 5:52 pm
- Forum: Formal Charge and Oxidation Numbers
- Topic: Stable Vs. Formal
- Replies: 4
- Views: 1107
Re: Stable Vs. Formal
Formal charge would refer to the charge of each individual atom and the overall number of electrons an atom owns in a molecule. I believe stable charge would just be the most stable form of the entire molecule.
- Tue Nov 13, 2018 5:35 pm
- Forum: Resonance Structures
- Topic: Resonance Structures
- Replies: 6
- Views: 860
Re: Resonance Structures
Resonance is a blending of structures. For example, a nitrate ion lewis structure depicts N as the central atom with bonds to three O atoms. Nitrate has one double bond and two single bonds. The double bond is connected to an oxygen, however, it can be connected to any of the three O atoms. This mak...
- Thu Nov 08, 2018 2:42 pm
- Forum: Ionic & Covalent Bonds
- Topic: Chemical formula from compound name
- Replies: 2
- Views: 440
Re: Chemical formula from compound name
The Roman numerals indicate the charge. For example, this means that Indium has a 3+ charge. We know that sulfur has a 2- charge, so the compound would be made of 2 Indium and 3 Sulfur to cancel out to charges. This results in the chemical formula: In2S3.
- Thu Nov 08, 2018 2:39 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: HW for discussion
- Replies: 9
- Views: 1066
Re: HW for discussion
The homework we turn in should correspond to what we've learned in class recently from the past week.
- Tue Nov 06, 2018 1:27 pm
- Forum: Octet Exceptions
- Topic: Central atoms with more than 8 electrons
- Replies: 3
- Views: 1755
Re: Central atoms with more than 8 electrons
Elements such as P or S do have open d-orbitals. This is because they have 3s and 3p orbitals already and have available 3d orbitals that can be filled next, allowing them to break the octet rule and have expanded valence shells. On the other hand, this wouldn't work for elements before period 3, be...
- Tue Oct 30, 2018 1:54 pm
- Forum: Formal Charge and Oxidation Numbers
- Topic: Stable Structure in SO4^-2
- Replies: 1
- Views: 692
Re: Stable Structure in SO4^-2
You might be mixing up FC with the overall charge of the ion. I think he said when the formal charge equals 0, that's when it's most stable. In both lewis structures he drew of the sulfate ion, they both resulted in having the -2 charge of the ion. In the first structure with single bonds, each oxyg...
- Tue Oct 30, 2018 12:09 am
- Forum: Ionic & Covalent Bonds
- Topic: 6th edition 3.9 and 3.11
- Replies: 1
- Views: 159
Re: 6th edition 3.9 and 3.11
These questions are basically asking which metal ions with a 2+ charge would have the electron configuration given? The 2+ charge means that 2 electrons were taken away to get the positive charge. For instance, 3.9a) [Ar]3d^7 Add back 2 electrons to get [Ar]3d^7, 4s^2. (Remember 4s fills up before 3...
- Tue Oct 30, 2018 12:03 am
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: How to write electron config for Bi3+?
- Replies: 1
- Views: 885
Re: How to write electron config for Bi3+?
Xenon's electron configuration is [Kr]4d^10, 5s^2, 5p^6. The ground state electron configuration for Bi is [Xe]4f^14, 5d^10, 6s^2, 6p^3 because you have to continue off that sequence. The 4f^14 and 5d^10 are from period 6 Lanthanoids and transition metals. It may not show as a single row on the PT, ...
- Wed Oct 24, 2018 10:18 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: #28 Atomic Spectroscopy Post-Assessment for module
- Replies: 1
- Views: 241
Re: #28 Atomic Spectroscopy Post-Assessment for module
What I did was that since it says "the meter" was defined as 1650763.73 wavelengths, I divided 1 meter / 1650763.73 wavelengths. I think this should get you one wavelength.
- Tue Oct 23, 2018 3:34 pm
- Forum: Properties of Light
- Topic: Is light in waves or photons?
- Replies: 10
- Views: 968
Re: Is light in waves or photons?
Light has both wave and photon properties. Wave properties of light explain the inverse relationship between wavelength and frequency of light. It also explains how amplitude represents the intensity of the wave. In this case, bigger amplitude (more intensity light) means greater energy. Meanwhile, ...
- Tue Oct 23, 2018 12:52 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: differences in equation
- Replies: 2
- Views: 339
Re: differences in equation
I think in the textbook the h has a line through it. This is a different h value that equals h/(2pi). That's why the h/2 is still equal to h/(4pi). The equation is the same; the notation of the h just makes it appear different.
- Tue Oct 23, 2018 11:53 am
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Series and Wavelengths part 2
- Replies: 1
- Views: 173
Re: Series and Wavelengths part 2
The Balmer series has lines with n(1) = 2 and n(2) > 2. It consists of lines that go from n = 2 as well as value anything above that as the second n value. So we know: 656.3 is n = 3 --> n = 2 486.1 is n = 4 --> n = 2 434.0 is n = 5 --> n = 2 410.2 is n = 6 --> n = 2 So now the next one would be : ?...
- Mon Oct 15, 2018 11:37 pm
- Forum: DeBroglie Equation
- Topic: Wavelength question 1.57 (6th edition)
- Replies: 1
- Views: 255
Re: Wavelength question 1.57 (6th edition)
So you know that the equation for the hydrogen spectrum is En = -hR/n^2. where R = 3.29E15 Hz. In this case you know that the Balmer series is where there is a transition from any energy level greater than 2 to n = 2. So 656.3 is n = 3 --> n = 2, 486.1 is n = 4 --> n = 2, 434.0 is n = 5 --> n = 2, 4...
- Mon Oct 15, 2018 7:04 pm
- Forum: Properties of Light
- Topic: When is the midterm?
- Replies: 1
- Views: 186
Re: When is the midterm?
The midterm is Monday, November 5, 6-8pm.
It's on his site under "test and exam schedule"
https://lavelle.chem.ucla.edu/wp-conten ... hedule.pdf
It's on his site under "test and exam schedule"
https://lavelle.chem.ucla.edu/wp-conten ... hedule.pdf
- Mon Oct 15, 2018 5:55 pm
- Forum: Photoelectric Effect
- Topic: HW Question 1.33
- Replies: 1
- Views: 254
Re: HW Question 1.33
Part c asks you to find the wavelength of the energy of the photon. In this case, you would use the equation E(photon) - work function = E(kinetic energy), where E(photon) = work function + E(kinetic energy). So from part b you should have found the work function to be 1.66E-17 J. Convert the given ...
- Mon Oct 15, 2018 5:41 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Question 1.15 sixth edition
- Replies: 10
- Views: 839
Re: Question 1.15 sixth edition
First, I think you mean that the frequency you got is 2.92E15 Hz (s^-1) On page 7 of Section 1.3, there is a figure that shows 102.6 nm is part of the Lyman series, which is on n = 1, so you know that the first energy level is n = 1. In order to find what it transitions to, you can use the equation ...
- Thu Oct 11, 2018 12:58 pm
- Forum: DeBroglie Equation
- Topic: Homework Question 1.37 6th Ed
- Replies: 2
- Views: 178
Re: Homework Question 1.37 6th Ed
I think this question is meant to show that their wavelengths are basically going to be the same. Since they are so close in size, there is very little difference, so you can say they are almost identical.
- Tue Oct 09, 2018 10:27 pm
- Forum: Photoelectric Effect
- Topic: Post-Mod Problem #28
- Replies: 4
- Views: 485
Re: Post-Mod Problem #28
The mass of an electron is always about 9.11E-31 kg.
- Tue Oct 09, 2018 5:40 pm
- Forum: Photoelectric Effect
- Topic: Work Function, Kinetic Energy
- Replies: 2
- Views: 656
Re: Work Function, Kinetic Energy
E(kinetic energy of e-) = .5mv^2.
The mass of the electron will usually be given to you.
The mass of an electron is about 9.11E-31 kg. Plug it in and you should get your answer!
The mass of the electron will usually be given to you.
The mass of an electron is about 9.11E-31 kg. Plug it in and you should get your answer!
- Tue Oct 09, 2018 5:33 pm
- Forum: Photoelectric Effect
- Topic: Post-Mod Problem #28
- Replies: 4
- Views: 485
Re: Post-Mod Problem #28
You know that E(photon) - E(energy to remove e-) = E(kinetic energy of e-). You also know that E(photon) = h(planck's constant) x v (frequency) In other words, hv - work function = (1/2)mv^2 (where m = mass of e- and v = velocity of e-) Mass of e- = 9.11E-31, Given velocity = 6.61E5 This would then ...
- Tue Oct 09, 2018 2:43 pm
- Forum: Photoelectric Effect
- Topic: Photoelectric Effect and Experiment
- Replies: 2
- Views: 172
Re: Photoelectric Effect and Experiment
In the photoelectric experiment, light was shone on a metal surface. If the energy of the incoming light was greater than the energy needed to remove the electron (work function), then electrons were emitted from the surface. This way, we are able to measure their kinetic energy.(Ephoton - Ework fun...
- Tue Oct 02, 2018 6:07 pm
- Forum: Molarity, Solutions, Dilutions
- Topic: Titrations v Dilutions
- Replies: 2
- Views: 1081
Re: Titrations v Dilutions
Titrations are when you measure the volume of one solution and see how much you need for it to react with the volume of another solution. For example, think of acid-base titrations. You know the amount of volume of the acid in a flask, but you don't know how much base is needed to react and neutrali...
- Tue Oct 02, 2018 5:52 pm
- Forum: Molarity, Solutions, Dilutions
- Topic: Find concentration of ions given more than one solute
- Replies: 2
- Views: 402
Re: Find concentration of ions given more than one solute
1. First you would determine the amount of moles for KCl, K2S, and K3PO4 by dividing the mass given by the molar mass of each compound. In that case, you would get 0.00671 mol KCl, 0.004535 mol K2S, and 0.00236 mol K3PO4. 2. In order to calculate the amount of potassium ions, you would take into acc...
- Tue Oct 02, 2018 5:32 pm
- Forum: Significant Figures
- Topic: Significant Figures
- Replies: 6
- Views: 808
Re: Significant Figures
When you do sig figs for addition/subtraction, only think about the amount of digits after the decimal point. First you would do addition/subtraction normally, then afterwards you would round your answer to the least number of places in the decimal portion of the numbers you added/subtracted. For ex...