Search found 77 matches

by 705192887
Sun Mar 17, 2019 12:44 am
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: adiabatic vs isothermal
Replies: 2
Views: 195

Re: adiabatic vs isothermal

adibiatic is when there's no heat transfer, and isothermal is when the temperature is constant/no change in T
by 705192887
Sat Mar 16, 2019 7:47 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: 15.65 C 6th edition
Replies: 1
Views: 42

Re: 15.65 C 6th edition

The reaction profile shows an endothermic reaction, so the reverse reaction has a lower activation barrier than the forward reaction. From this, we know that the rate constant of the reaction with the higher activation barrier will increase when temperature is raised. Since the reaction is endotherm...
by 705192887
Sat Mar 16, 2019 7:38 pm
Forum: Administrative Questions and Class Announcements
Topic: LYNDON'S PORK RAMEN REVIEW
Replies: 37
Views: 2588

Re: LYNDON'S PORK RAMEN REVIEW

Can someone who went to the review session post pictures of it? Thanks
by 705192887
Sat Mar 16, 2019 12:06 am
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: Collisions/Temperature dependence
Replies: 2
Views: 57

Collisions/Temperature dependence

Can anyone explain this, it's from the Kinetics outline:

Explain how the collision model and activated complex model account for the temperature
dependence of reactions.
by 705192887
Fri Mar 15, 2019 9:10 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Relationship b/w ∆H and ∆S
Replies: 1
Views: 187

Relationship b/w ∆H and ∆S

So on the Thermodynamics outline on the class website, one of the bullet points says: "Show how ∆S is related to ∆H for a change at constant temperature and pressure and explain the relationship"

Can someone explain this concept and how to show it?
by 705192887
Fri Mar 15, 2019 7:21 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: Question 15.73 (Sixth Edition)
Replies: 1
Views: 42

Re: Question 15.73 (Sixth Edition)

A is false because catalysts never change the equilibrium constant. It changes both the rate of the forward and reverse reactions B is true because a catalyst is a substance that increases the rate of reaction, but is not consumed in it. C is false because a catalyst provides a new pathway for the r...
by 705192887
Fri Mar 15, 2019 6:42 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: addition of solid into a reaction
Replies: 1
Views: 40

Re: addition of solid into a reaction

The reaction will not shift either way because solids are not included in the equilibrium constant. Thus, adding a solid will not affect the Q value nor the reaction.
by 705192887
Fri Mar 15, 2019 6:40 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 14.37
Replies: 1
Views: 41

Re: 14.37

The anode and the cathode are the same in this cell diagram, so the E°cell = 0. You dont need the specific value for HCl because it's the same on both sides. Now just use the Nernst equation to find Ecell from the difference in concentrations.
by 705192887
Thu Mar 14, 2019 12:43 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: pH question
Replies: 1
Views: 57

Re: pH question

First, write out the balanced reaction: C7H5O2 + water --> C7H6O2 + OH-. This reaction is basic, so you need to convert pKa into pKb to use for this reaction. 14-pKa=pKb, so pKb = 9.596. Next, convert pKb into Kb in order to use it in an ice table. 10^-pKb = Kb = 2.535x10^10. Now, from an ice table,...
by 705192887
Thu Mar 14, 2019 12:16 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: comparing forward and reverse rates
Replies: 4
Views: 73

Re: comparing forward and reverse rates

The rate constants themselves are not the same for forward and reverse reactions at equilibrium. The rates are equal, but not their K values.
by 705192887
Thu Mar 14, 2019 12:14 pm
Forum: Second Order Reactions
Topic: Question 15.19 on 6th edition hw
Replies: 1
Views: 45

Re: Question 15.19 on 6th edition hw

So to find the reaction of B, you divide reaction 2 by reaction 4. This cancels out K, A, and C, leaving just the rate in relation to B. (8.7/50.8)=(1.25)^y/(3.02)^y. By solving this equation, you get (1/5.8)=(1/2.4)^y. This simplifies to (2.4)^2=(2.4)^y. Therefore, y, which is the exponent on [B] a...
by 705192887
Tue Mar 05, 2019 4:50 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: Q: Partial Pressures and Concentration?
Replies: 4
Views: 110

Re: Q: Partial Pressures and Concentration?

I believe that since they give one part in pressure and one in concentration, you can just use those values to find E=E°-(.025693V/n)ln(P/[M]). This still gives a usable relationship between the products and reactants. One half as 1 torr pressure with different concentrations while the other half ha...
by 705192887
Wed Feb 27, 2019 4:45 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 14.11 6th Edition
Replies: 1
Views: 52

14.11 6th Edition

Can someone help me with 14.11 in the 6th edition? I'm studying for the test and practicing problems but the solutions manual seems to be backwards from what I'm expecting every time...
by 705192887
Tue Feb 26, 2019 11:39 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Week 6 Friday Lecture Notes
Replies: 2
Views: 58

Week 6 Friday Lecture Notes

I was sick on Friday of week 6 (2/15) and was wondering if anyone could post a picture of their notes? Thanks!
by 705192887
Tue Feb 26, 2019 5:01 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Entropy and Enthalpy
Replies: 2
Views: 68

Re: Entropy and Enthalpy

Enthalpy and entropy are somewhat related to each other. If energy is added to or released from the system, it has to be partitioned into new states. Therefore an enthalpy change can also have an effect on entropy.
by 705192887
Tue Feb 26, 2019 12:27 pm
Forum: Balancing Redox Reactions
Topic: HOMEWORK 14.5 D
Replies: 4
Views: 58

Re: HOMEWORK 14.5 D

Will someone explain how to balance d in chapter 14 problem 5 in the 6th edition. I don't understand why the OH is on the right side instead of the left when you would add H2O on the left and therefore OH- on the right First, you want to balance the oxygen by adding 8 H2O molecules on the left. Now...
by 705192887
Wed Feb 20, 2019 5:16 pm
Forum: Balancing Redox Reactions
Topic: 7th edition, 6K3 pt d
Replies: 1
Views: 39

Re: 7th edition, 6K3 pt d

For this reaction, I think that the Cl2 is being oxidized and also being reduced. It's oxidation number goes up from O to +1 in HClO and it goes down from 0 to -1 in Cl-.
by 705192887
Wed Feb 20, 2019 5:06 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: Gibb's free energy Units
Replies: 4
Views: 70

Re: Gibb's free energy Units

Depending on what the units are in the problem, you can give your answer in terms of J or KJ as they are essentially interchangeable. Normally, it will be in terms of KJ/mol I presume.
by 705192887
Mon Feb 18, 2019 9:12 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Gibbs Free Energy
Replies: 4
Views: 76

Re: Gibbs Free Energy

Yes. Here, K is the equilibrium constant, which is [products]/[reactants] at equilibrium
by 705192887
Wed Feb 13, 2019 12:51 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Delta S equations
Replies: 3
Views: 113

Re: Delta S equations

Hi Katherine! dS = q/t when you're dealing with a reversible process. dS = nRln(Vf/Vi) when the process is isothermal. dS = nCln(Tf/Ti) when there's a change in temperature associated with the change in entropy. Your C value will change based on whether your volume (Cv) or pressure (Cp) are constan...
by 705192887
Tue Feb 12, 2019 3:03 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Help on 9.5
Replies: 3
Views: 66

Re: Help on 9.5

Because heat is lost by one and absorbed by the other. Negative for one and positive for the other.
by 705192887
Tue Feb 12, 2019 2:44 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Example 8.3 sixth edition
Replies: 3
Views: 86

Re: Example 8.3 sixth edition

They don't need to convert it because Celsius is Just kelvin—273. Therefore an increase of 80°C is also an increase of 80°K.
∆T(°C) = ∆T(K)
by 705192887
Tue Feb 12, 2019 1:50 pm
Forum: Administrative Questions and Class Announcements
Topic: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!
Replies: 49
Views: 2947

Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

If he said he isn't gonna post the answers, could someone post them so I can check my answers?
by 705192887
Sun Feb 10, 2019 6:25 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Hmwrk 8.67 6th edition
Replies: 1
Views: 70

Re: Hmwrk 8.67 6th edition

For part A, you want to find the enthalpy of the reaction H2g + (.5)O2g ——> H2O(l) Now use bond enthalpies to find the enthalpy change: 1mol(436KJ/mol) to break the H-H bonds in H2 and 1mol(496KJ/mol) for the O-O bonds in O2. This is the breaking of the reactant bonds 2mol(463KJ/mol) to form the 2mo...
by 705192887
Sun Feb 10, 2019 6:13 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: 6th edition 9.15; standard conditions? (Degree symbol)
Replies: 1
Views: 57

Re: 6th edition 9.15; standard conditions? (Degree symbol)

I believe that it might be because part a is given exactly one mole of the substance, so that's standard conditions? While part b gives a mass in grams of the substance which then must be converted to moles. This is the only difference that I can think of, other than maybe that in part a it's at 0°C...
by 705192887
Sun Feb 10, 2019 6:09 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Homework: 6h edition, 12.55
Replies: 1
Views: 59

Re: Homework: 6h edition, 12.55

The pH difference according to molecular structure stems from the electron-withdrawing ability of the group attached to the acid. In other words, as electronegativity of the function group attached to the acid increases, the acidity of the acid will increase. Acidity therefore increases, let's say, ...
by 705192887
Sat Feb 09, 2019 6:13 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Positive or Negative Entropy Change
Replies: 2
Views: 63

Re: Positive or Negative Entropy Change

Yes because if the system gives off heat, they are losing internal energy and there is a decrease in entropy (negative delta S). Vice versa for the surroundings.
by 705192887
Thu Feb 07, 2019 11:48 am
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: 6th ed 8.9
Replies: 1
Views: 51

Re: 6th ed 8.9

Work is found in Joules, so I got -1.50x10^2 J for my work, which is essentially the same as your 148 value. However, I think you may be misinterpreting your 148 J value for 148 KJ. The answer I got was -1.50x10^2 J or -.150 KJ. Hope this helps! If not, I can run you through the steps of the problem.
by 705192887
Thu Feb 07, 2019 9:52 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 7th Edition 4.15
Replies: 2
Views: 82

Re: 7th Edition 4.15

Start out by balancing the equation: 2HCl + Zn --> H2 + ZnCl2 Find the limiting reactant (Zinc). Now use Hess's law to find the Heat of the reaction (ΔHrxn), which is just the sum of the heats of formation of products minus the sum of the heats of formation of the reactants. This should give you the...
by 705192887
Thu Feb 07, 2019 9:47 am
Forum: Administrative Questions and Class Announcements
Topic: Week 5 Homework
Replies: 3
Views: 93

Re: Week 5 Homework

Yes. At this point, I feel like anything we've done so far should suffice for HW, since it's all preparation for the midterm anyways.
by 705192887
Thu Feb 07, 2019 9:46 am
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Work on an adiabatic process?
Replies: 2
Views: 59

Re: Work on an adiabatic process?

Yes, work can still be done on an adiabatic system (no transfer of heat (q=0)). Delta U in this system will then just equal W since the value for q is 0. Hope this helps!
by 705192887
Thu Feb 07, 2019 9:43 am
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Help on 9.5
Replies: 3
Views: 66

Re: Help on 9.5

Entropy has the units energy over temperature, or joules per kelvin. You are correct that you need to add them together. However, you need to find the individual entropies by dividing the joules by their temperature in kelvin. Then add those values like the book to find entropy. hope this helps.
by 705192887
Wed Jan 30, 2019 10:23 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Problem 8.11 6th Edition
Replies: 1
Views: 37

Re: Problem 8.11 6th Edition

In general, I believe that the reversed reaction can do more work because less of the energy is lost as heat.
by 705192887
Sun Jan 27, 2019 10:23 pm
Forum: Phase Changes & Related Calculations
Topic: Steam
Replies: 11
Views: 200

Re: Steam

Yes. Steam has more energy so it causes more severe burns.
by 705192887
Sun Jan 27, 2019 9:28 pm
Forum: Phase Changes & Related Calculations
Topic: State Properites
Replies: 7
Views: 123

Re: State Properites

They are not considered state properties because their value depends on the path taken from the initial to final values. State properties do not depend on the path taken, such as pressure or temperature.
by 705192887
Wed Jan 23, 2019 4:49 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: ICE Table Clarification
Replies: 4
Views: 67

ICE Table Clarification

When using an ICE table to determine equilibrium concentrations, when are we allowed to assume x is negligible for the reactant concentration so that we don't have to deal with the quadratic? Thanks
by 705192887
Sun Jan 20, 2019 2:11 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 6th Edition Hw #12.79 using Strong Acids
Replies: 1
Views: 46

Re: 6th Edition Hw #12.79 using Strong Acids

The H3O+ concentration will be .15 M in the E column because the .15 M of H2SO4 dissociates completely into H3O+. The initial concentration of H3O+ will be 0 M. Hope this helps.
by 705192887
Wed Jan 16, 2019 5:08 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Acids
Replies: 6
Views: 91

Re: Acids

Strong acids and bases disassociate/ionize completely while weak acids and bases do not completely. this is the main difference.
by 705192887
Wed Jan 16, 2019 5:05 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Solids and liquids in K
Replies: 6
Views: 131

Re: Solids and liquids in K

Since there is excess water its concentration changes slightly and isn't taken into account since it's the solvent.
by 705192887
Sun Jan 13, 2019 8:30 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Catalyst
Replies: 4
Views: 48

Re: Catalyst

a catalyst only speeds up a reaction but doesn't actually change the k value. the only thing that can change the value of k/equilibrium constant is a change in temperature.
by 705192887
Sun Jan 13, 2019 8:27 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Equilibrium 1A module
Replies: 2
Views: 48

Re: Equilibrium 1A module

This is talking about whether the equilibrium sits more toward the product or the reactant side of the reaction. Sitting to the left refers to having a higher reactant concentration at equilibrium and K value that's less than 1. On the other hand, sitting to the right refers to having a higher produ...
by 705192887
Sun Jan 13, 2019 8:18 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: equilibrium
Replies: 4
Views: 60

Re: equilibrium

This is talking about whether the equilibrium sits more toward the product or the reactant side of the reaction. Sitting to the left refers to having a higher reactant concentration at equilibrium and K value that's less than 1. On the other hand, sitting to the right refers to having a higher produ...
by 705192887
Wed Dec 05, 2018 9:33 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: Hydrogen Bonds
Replies: 14
Views: 228

Re: Hydrogen Bonds

Yes, hydrogen bonds are a dipole-dipole interaction, and happen with a hydrogen and either O,N, or F. They are stronger than basic dipole-dipole interactions.
by 705192887
Wed Dec 05, 2018 9:26 am
Forum: Empirical & Molecular Formulas
Topic: midterm question
Replies: 3
Views: 239

Re: midterm question

The third "hidden" reactant is oxygen. It is in excess so its original mass is not given. Therefore, this excess oxygen can account for the discrepancy in mass between the products and reactants.
by 705192887
Wed Dec 05, 2018 9:24 am
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: unit
Replies: 3
Views: 61

Re: unit

Concentration should be in Molarity M, which is moles per liter (moles L^-1)
by 705192887
Wed Dec 05, 2018 9:22 am
Forum: Lewis Acids & Bases
Topic: Identifying the acid and base
Replies: 3
Views: 99

Re: Identifying the acid and base

In the equations, the acids are the compounds that donate/lose a hydrogen H+. Therefore, the acids are:
a)H3O+
b) CH3COOH
c) HI
by 705192887
Sun Dec 02, 2018 11:47 pm
Forum: Conjugate Acids & Bases
Topic: Clarification of Conjugate Acids and Bases
Replies: 4
Views: 104

Re: Clarification of Conjugate Acids and Bases

Conjugate acids and bases are molecules that differ by one proton. The conjugate acid is the molecule in the reaction that is formed by the acceptance of an H+ that is donated by the base. Example: Conjugate acid = H2O; Base = OH-. In the same way, conjugate bases and acids are similar in that they ...
by 705192887
Sun Dec 02, 2018 11:35 pm
Forum: Bronsted Acids & Bases
Topic: Strong Bases
Replies: 3
Views: 97

Re: Strong Bases

I believe that strong bases ionize completely in water because they are better proton donors. This means that they completely donate a proton while a weaker base does donate, but not as completely.
by 705192887
Sun Dec 02, 2018 11:29 pm
Forum: Hybridization
Topic: Hybridization
Replies: 11
Views: 174

Re: Hybridization

trigonal planar is sp2 since it has 3 electron densities, while tetrahedral would be sp3 since it has 4 electron densities.
by 705192887
Sun Nov 25, 2018 8:25 pm
Forum: Hybridization
Topic: Octet Expansion
Replies: 6
Views: 112

Re: Octet Expansion

The d orbital is needed because it has an extra 10 electron capacity. s and p are filled up first and can only hold 8 (full octet), so the extra 10 from d orbital are used to exceed the octet.
by 705192887
Sun Nov 25, 2018 8:22 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Sigma Bond vs. Pi Bond
Replies: 3
Views: 95

Re: Sigma Bond vs. Pi Bond

Pi bonds are fixed because there is an electron overlap below and on top of the atoms' plane. On the other hand, sigma bonds can be rotated because there is just an overlap of two orbitals.
by 705192887
Sun Nov 25, 2018 8:20 pm
Forum: Hybridization
Topic: Polar vs Non Polar
Replies: 3
Views: 73

Re: Polar vs Non Polar

Polarity of a molecule depends on whether or not the molecule is symmetrical in regards to its partial charges and sharing of electrons. If the molecule is symmetrical and has no dipole moments, or the dipole moments on opposite sides cancel out (like in water) the molecule is non-polar. However, if...
by 705192887
Mon Nov 19, 2018 9:34 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Double bonds
Replies: 8
Views: 171

Re: Double bonds

VSEPR does not take in account what type of bond a bond is. In this sense, a single bond affects VSEPR in the same way as a double bond and triple bond and so on. For VSEPR, a bond is a bond.
by 705192887
Mon Nov 19, 2018 9:19 pm
Forum: Bond Lengths & Energies
Topic: double bonds vs. single bonds
Replies: 5
Views: 132

Re: double bonds vs. single bonds

Double bonds are more stable than single bonds. This is because double bonds also have a π (Pi) bond, while single bonds only have σ (sigma) bonds. Pi bonds prevent rotation thus making double bonds more stable.
by 705192887
Mon Nov 19, 2018 9:00 pm
Forum: Bond Lengths & Energies
Topic: VSEPR
Replies: 6
Views: 183

Re: VSEPR

After completing the Lewis Dot structure, GeF4 has 4 single bonds and no lone pairs around the central atom. Therefore, from VSEPR, the shape is tetrahedral.
by 705192887
Mon Nov 12, 2018 2:36 pm
Forum: Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)
Topic: Hydrogen Bonds
Replies: 14
Views: 228

Re: Hydrogen Bonds

Yes, Hydrogen Bonds are dipole-dipoles with N,O, and F. Not all dipole-dipoles are hydrogen bonds though, while all hydrogen bonds are dipole dipoles.
by 705192887
Tue Nov 06, 2018 4:21 pm
Forum: Formal Charge and Oxidation Numbers
Topic: Formal Charge of Oxygen
Replies: 6
Views: 160

Re: Formal Charge of Oxygen

In general, the more electronegative atom should have the negative formal charge since the electrons are more strongly attracted/pulled to them.
by 705192887
Tue Nov 06, 2018 4:20 pm
Forum: Bond Lengths & Energies
Topic: Bond Lengths
Replies: 6
Views: 151

Re: Bond Lengths

Bond length decreases as the number of bonds increases since the two atoms are more strongly attracted to one another. For example, triple bonds are shorter than double bonds which are shorter than single bonds. When determining bond length, take the average of all the bond lengths in the resonance ...
by 705192887
Thu Nov 01, 2018 10:03 pm
Forum: Administrative Questions and Class Announcements
Topic: MIDTERM PRACTICE - Garlic Bread Review Session [ENDORSED]
Replies: 121
Views: 5452

Re: MIDTERM PRACTICE - Garlic Bread Review Session [ENDORSED]

Can anyone help me with 9b? It asks for the electron configuration of Chromium. I got an answer of [Ar]4s2 3d4 but I looked it up online and it says that its' actually [Ar]4s1 3d5. Why is one of the 4s electrons actually in the 3d orbital? I looked it up and it mentioned something about stability? S...
by 705192887
Thu Nov 01, 2018 4:15 pm
Forum: Ionic & Covalent Bonds
Topic: 6th edition, 3.11
Replies: 1
Views: 24

Re: 6th edition, 3.11

This question is asking what metal ion with a positive 3 charge (3+) will have the following electron configurations. This confused me at first too, but M is not a symbol for an element, but rather a placeholder/variable for any metallic element. After realizing that, I began writing out the electro...
by 705192887
Thu Nov 01, 2018 4:06 pm
Forum: Lewis Structures
Topic: Most stable lewis
Replies: 3
Views: 93

Re: Most stable lewis

Formal charge is used to determine the stability of a molecule. However, resonance structures, since they are different arrangements of the same bonds, are equally stable as one another.
by 705192887
Thu Nov 01, 2018 4:04 pm
Forum: DeBroglie Equation
Topic: De Broglie vs. speed of light/Einstein equation
Replies: 3
Views: 238

Re: De Broglie vs. speed of light/Einstein equation

De Broglie's wavelength equation is used for particles and objects, such as electrons or a baseball, while the speed of light/Einstein equation is used only when calculating the wavelength or frequency of light. C=λv can't be used with particles because it describes a relationship with light. This i...
by 705192887
Thu Nov 01, 2018 4:00 pm
Forum: Lewis Structures
Topic: Lewis Structure NO3-
Replies: 3
Views: 62

Re: Lewis Structure NO3-

Nitrogen has 5 valence electrons, so it needs three more electrons to fulfill its octet. This doesn't necessarily mean that it's going to make 3 bonds, but rather it starts with 5 electrons when determining the lewis structure. It will make its bonds to fill its octet.
by 705192887
Tue Oct 30, 2018 3:00 pm
Forum: General Science Questions
Topic: Midterm studying
Replies: 10
Views: 387

Re: Midterm studying

I found that completing the online modules on the class website is helpful in reviewing lecture material and building a strong conceptual foundation. Then, I simply do many hw problems to master the different types of problems. I also plan on going to some review sessions this week.
by 705192887
Tue Oct 30, 2018 2:57 pm
Forum: Resonance Structures
Topic: Lengths of bonds
Replies: 3
Views: 97

Re: Lengths of bonds

Double Bonds are shorter because they share more electrons between the two atoms, meaning that those two atoms are more tightly bonded are therefore closer together. The attraction between the two is greater than if it were just a single bond. Similarly, a triple bond would then have a shorter bond ...
by 705192887
Tue Oct 30, 2018 2:53 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: How to do e- configuration shorthand?
Replies: 3
Views: 88

Re: How to do e- configuration shorthand?

You would just use [Xe] since it has the same electron configuration as that noble gas. No need to go backwards to [Kr].
by 705192887
Sun Oct 28, 2018 11:45 pm
Forum: Ionic & Covalent Bonds
Topic: Roman numerals next to element
Replies: 8
Views: 579

Re: Roman numerals next to element

The roman numeral signifies what the charge on that atom is. For example, depending on what Iron or another element is bonded to, it could be Iron II or Iron III to balance the charges. Hope this helps! #sickomode
by 705192887
Sun Oct 28, 2018 11:43 pm
Forum: Ionic & Covalent Bonds
Topic: Question 2B.3 part d
Replies: 2
Views: 60

Re: Question 2B.3 part d

To my knowledge, this is simply an exception to the octet rule.
by 705192887
Sun Oct 28, 2018 11:41 pm
Forum: Ionic & Covalent Bonds
Topic: Covalent Bonds
Replies: 16
Views: 325

Re: Covalent Bonds

No such thing as a dumb question, David!! And, to answer your question, nonmetals (right side of periodic table) do not form cations because they only gain a few electrons to fill their octet, meaning they become anions. Metals, on the other hand, lose electrons to fill their octet. Hope this helps!...
by 705192887
Sun Oct 28, 2018 11:31 pm
Forum: Trends in The Periodic Table
Topic: Periodic Trend - Ionization energy
Replies: 2
Views: 52

Re: Periodic Trend - Ionization energy

You are correct, Fluorine has a higher ionization energy than Chlorine. However, I'm pretty sure that Helium has the highest first ionization energy because it's outermost electron is closest to the nucleus.
by 705192887
Thu Oct 11, 2018 1:02 am
Forum: Molarity, Solutions, Dilutions
Topic: Post Module Assessment
Replies: 2
Views: 309

Post Module Assessment

So I'm taking the first test tomorrow and I am slightly confused on one of the post-module assessment questions in the Molarity and Dilution section. I was wondering if anyone could just run me through as to how to go about solving this problem? I know to start by finding the initial concentration o...
by 705192887
Thu Oct 11, 2018 12:56 am
Forum: Empirical & Molecular Formulas
Topic: Question M19: Empirical and Molecular formula of caffeine [ENDORSED]
Replies: 4
Views: 193

Re: Question M19: Empirical and Molecular formula of caffeine [ENDORSED]

By process of elimination, you know that the hydrogen, carbon, and nitrogen must come from the caffeine molecule, since they couldn't have come from O2. After this, solve for the empirical formula using the mass of the products formed, and those masses are the same as the masses in the original caff...
by 705192887
Thu Oct 11, 2018 12:47 am
Forum: DeBroglie Equation
Topic: Homework Question 1.37 6th Ed
Replies: 2
Views: 48

Re: Homework Question 1.37 6th Ed

This is just a guess, but maybe use the mass for the neutron as the mass from the isotope? That way, the two masses are slightly different and will have different wavelengths. For example, carbon has an atomic mass of 12.011. The proton weight would be 6, and the neutron weight would be 6.011. Maybe...
by 705192887
Mon Oct 08, 2018 12:00 am
Forum: Molarity, Solutions, Dilutions
Topic: Molarity formula
Replies: 6
Views: 253

Re: Molarity formula

Yes, the first is essentially derived from the latter. Sometimes you will need to use just the M1V1=M2V2, depending on what is given, but sometimes you will need both to complete the problem.
by 705192887
Sun Oct 07, 2018 11:46 pm
Forum: Molarity, Solutions, Dilutions
Topic: Actual/Theoretical Yield
Replies: 8
Views: 1051

Re: Actual/Theoretical Yield

I doubt that specifics are needed as to why the actual yield is less than the theoretical yield, especially since this is a "relatively" introductory chemistry class. However, it is good to know some possibilities or why in general this disparity occurs. For example, since the theoretical ...
by 705192887
Sun Oct 07, 2018 11:35 pm
Forum: Limiting Reactant Calculations
Topic: Question M3
Replies: 3
Views: 89

Re: Question M3

Normally, yes. Both masses should be given to then determine which reactant is limiting and which is in excess. If only one is given, I would assume that that reactant is the limiting reactant since no other information is given. However, based on the context of the question, you may want to always ...

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