## Search found 59 matches

Wed Mar 13, 2019 5:10 am
Forum: Second Order Reactions
Topic: Question 15.13 Part B (Sixth Edition)
Replies: 2
Views: 140

### Re: Question 15.13 Part B (Sixth Edition)

This is not a question that requires calculation as much as it is a question that requires conceptual understanding. In a 1st order reaction, the rate of the reaction varies proportionally with the concentration of reactant. But in a 2nd order reaction, the rate will quadruple if the concentration o...
Wed Mar 13, 2019 5:08 am
Forum: Second Order Reactions
Topic: Final
Replies: 30
Views: 790

### Re: Final

The final will definitely cover all of the material that we have learned over the course of this year. You can expect conceptual questions as well as questions that require calculations. I think that most of the material on the final will be on the material that we covered after the midterm, so expe...
Wed Mar 13, 2019 5:04 am
Forum: Second Order Reactions
Topic: 7B.13 Help
Replies: 4
Views: 176

### Re: 7B.13 Help

For many of these types of problems, I think you should first understand the equations that you would use. Once you know which equations are available, you can maneuver the date given to find the answer you need. For this problem specifically, you need the half life equation and the equation that re...
Tue Mar 05, 2019 2:55 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: n and k
Replies: 3
Views: 149

### Re: n and k

You need to find n to find what order that reactant is. The order will determine by what factor the change in concentration of that reactant affects the overall rate of reaction. You need to find k because that is the rate constant that is specific to that reaction that you need to find the rate of ...
Tue Mar 05, 2019 2:53 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: Which Rate to Use When Finding K? HW Chapter 15 #19, 6th Edition
Replies: 2
Views: 121

### Re: Which Rate to Use When Finding K? HW Chapter 15 #19, 6th Edition

It doesn't matter which experiment you use because k should be the same. You should definitely check again to see if the k values are different for every experiment because that shouldn't be the case. As long as you are consistent with the values of a certain reaction, then you should be good.
Tue Mar 05, 2019 2:52 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: K
Replies: 2
Views: 153

### Re: K

You will use any one of the experiments to determine k. Assuming you have the correct rate law for the reaction, you simply plug in the rate, concentrations of the reactants, and solve for k.
Tue Mar 05, 2019 2:51 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: Zero, First, and Second Order reactions
Replies: 2
Views: 135

### Re: Zero, First, and Second Order reactions

Instead of looking at the molar coefficients in the reaction, you would only use the exponents of the reactants to determine the overall order of the reaction. So, when you write the rate law of the reaction, H2 and I2 will each have an order of 1 which you add up to get an overall order of 2.
Tue Mar 05, 2019 2:49 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: Changing concentration to change rate
Replies: 2
Views: 138

### Re: Changing concentration to change rate

If a reactant is second-order, this means that doubling that reactant will result in the rate being quadrupled. This is because the concentration of the reactant is to the second power. Therefore, if you double the concentration (multiply by 2), then the 2 is squared to give you rate that is quadrup...
Tue Mar 05, 2019 2:47 pm
Forum: General Rate Laws
Topic: 6th edition 15.9
Replies: 4
Views: 172

### Re: 6th edition 15.9

In general, the rule of the thumb for finding the units of k is to find what units will result in the rate having the units of mol/(L*s). First, look at the units that the concentrations give you and make the units of k cancel those concentrations leaving you with mol/(L*s).
Tue Mar 05, 2019 2:45 pm
Forum: General Rate Laws
Topic: 15.9 6th edition
Replies: 3
Views: 138

### Re: 15.9 6th edition

Normally, you will see that changing the concentration of the reactants will affect the rate of the reaction. However in a zero-order reaction, changing the concentration will not change the rate. Therefore, the rate will equal the k constant. In order for the rate to have the units of mol/(L*s), k ...
Mon Feb 25, 2019 1:52 am
Forum: Balancing Redox Reactions
Topic: Salt Bridges-won't they dissolve?
Replies: 5
Views: 198

### Re: Salt Bridges-won't they dissolve?

LOL. The salt bridge provides no other purpose than to allow the free flow of ions from each side of the cell to prevent the polarization of one side that would stop the reaction from occurring. It's not made of salt so it doesn't change the concentration of either solution.
Mon Feb 25, 2019 1:48 am
Forum: Balancing Redox Reactions
Topic: 6K3(d) 7th edition
Replies: 1
Views: 100

### Re: 6K3(d) 7th edition

When Cl2 is broken up into the constituent Cl ions, then one will obviously react with water to create HClO. The remaining Cl will react with the H+ ion leftover from the creation of HClO to form HCl. But since HCl is a strong acid, it immediately dissociates into H+ and Cl-. This, I believe, is why...
Mon Feb 25, 2019 1:44 am
Forum: Balancing Redox Reactions
Topic: Rules for redox reactions
Replies: 3
Views: 119

### Re: Rules for redox reactions

I think you just have to figure out the charges on your own based off of other information. In the example Lavelle gave in class, he used the reaction between MnO4- and Fe+2. On the products side of the reaction, you see that MnO4- has a net charge of -1, and since you know that O has a charge of -2...
Tue Feb 19, 2019 11:57 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Enthalpy of phase change
Replies: 5
Views: 384

### Re: Enthalpy of phase change

There are two parts to the process in which Br2 becomes 2Br. One part is the breaking of the bond between Br2, and the second part is the vaporization of Br2 to a gas.
Tue Feb 19, 2019 11:54 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Determine if molecule is in most stable form
Replies: 5
Views: 360

### Re: Determine if molecule is in most stable form

I think you should just draw the Lewis structure of the molecule and check to see if the formal charges of the atoms are all 0 or close to 0.
Tue Feb 19, 2019 11:53 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: HW Problem 9.5
Replies: 2
Views: 149

### Re: HW Problem 9.5

A lot of these problems involve adding up the steps similar to how you do a Hess's Law problem.
Sat Feb 09, 2019 11:45 pm
Forum: Phase Changes & Related Calculations
Topic: 4C.13 7th Ed.
Replies: 2
Views: 105

### Re: 4C.13 7th Ed.

If you look at josephyim1H's post in this same thread you will see an explanation for this problem. Basically, the gist of what I said was that the ice will first melt (use delta H) and then increase in temperature to the exact same temperature as the rest of the water that was already in the glass....
Sat Feb 09, 2019 11:41 pm
Forum: Phase Changes & Related Calculations
Topic: heat of a phase change
Replies: 5
Views: 204

### Re: heat of a phase change

For phase changes, keep in mind that the temperature of the substance in question does not change. You can see that if you look at a heating or cooling curve. Therefore, you can't use q=(m)(C)(delta T) because you would end up with 0 due to the net change in temperature being 0. Instead, you have to...
Sat Feb 09, 2019 11:38 pm
Forum: Phase Changes & Related Calculations
Topic: Example 8.41
Replies: 1
Views: 107

### Re: Example 8.41

The key to this problem is knowing that the heat gained by the water is equal to the heat lost by the water. If you think about what will happen to the system logically, then the ice will melt into water while simultaneously cooling down the water already in the glass. All that's left in the glass w...
Sun Feb 03, 2019 10:33 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Hess's Law
Replies: 3
Views: 143

### Re: Hess's Law

I doubt it because the numbers that we use in the formula usually are at the same temperature.
Sun Feb 03, 2019 10:30 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Delta H and q
Replies: 3
Views: 104

### Re: Delta H and q

I'm pretty sure q and delta H are the same thing.
Sun Feb 03, 2019 9:48 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Bomb Calorimeter
Replies: 5
Views: 240

### Bomb Calorimeter

Can someone explain what a bomb calorimeter does? Thanks in advance.
Sun Jan 27, 2019 4:41 pm
Forum: Ideal Gases
Topic: SI Units
Replies: 5
Views: 367

### Re: SI Units

We will usually use atm, but other common units for pressure include kPa, torr, and mm Hg.
Sun Jan 27, 2019 4:39 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Changing Pressure
Replies: 6
Views: 290

### Re: Changing Pressure

Inert gases do not react with the reactants nor the products, so it doesn't really have any effect on the concentrations of either. However, when you change the volume, then you can figure out which reaction will be favored by looking at the number of moles on both sides of the reaction.
Sun Jan 27, 2019 4:38 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Replies: 10
Views: 806

The concentrations of the reactants and the products both don't change when you add an inert gas because inert gases don't react with anything really.
Mon Jan 21, 2019 8:09 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: 5J13 7th Ed
Replies: 2
Views: 108

### Re: 5J13 7th Ed

You're confusing exothermic/endothermic with a mere change in temperature. Just because they heat the mixture from 600 to 700 doesn't tell you anything about the reaction being exothermic/endothermic. Therefore, you can answer the question by looking at the k values that give you for each temperatur...
Mon Jan 21, 2019 8:01 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: le chatliers
Replies: 3
Views: 124

### Re: le chatliers

Notice that C is a solid. Therefore the concentration of C won't change. Then, if you look at the equation again, there are more moles of gas on the products side than the reactants. This is why the reaction will favor the reactants side if you increase the pressure.
Mon Jan 21, 2019 7:56 pm
Forum: Equilibrium Constants & Calculating Concentrations
Replies: 2
Views: 64

Typically, you use an ICE table when the question asks for the equilibrium concentrations of a reaction and they give you the initial concentrations. Then when you construct your ICE table, you will fill in the blanks for the initial concentrations but use "x" for the change in concentrati...
Sat Jan 12, 2019 11:44 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 11.43 in 6th Edition
Replies: 2
Views: 62

### Re: 11.43 in 6th Edition

You should probably convert one of them to the other. And yes, it's usually safe to assume the starting concentration/partial pressures of the products are 0 if not explicitly stated.
Sat Jan 12, 2019 11:41 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Partial Pressure and Concentration [ENDORSED]
Replies: 3
Views: 105

### Re: Partial Pressure and Concentration[ENDORSED]

Using PV=nRT, you can rearrange the expression to read P=(n/V)RT. The (n/V) is essentially concentration so you can convert from partial pressure to concentration this way.
Sat Jan 12, 2019 11:38 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Question 11.11 (Sixth Edition)
Replies: 1
Views: 73

### Re: Question 11.11 (Sixth Edition)

a/b) If you start with more moles of O3 then the partial pressure of O3 will increase. You know that the Kp must remain constant so, looking at the equation for Kp = (PO2)^3/(PO3)^2 you can reason that the partial pressure of O2 will also increase to keep the ratio the same. The only way that the pa...
Sun Dec 02, 2018 10:10 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: repulsion
Replies: 3
Views: 183

### Re: repulsion

The electrons in a lone pair have more energy than bonded electrons because they are free and not restricted to a bond. Therefore, they repel each other more because they are more excited and take up more space. This is why they decrease bond angles and slightly distort the molecular geometry of mol...
Sun Dec 02, 2018 10:01 pm
Forum: Hybridization
Topic: hybridization
Replies: 13
Views: 659

### Re: hybridization

Like other people said before, count the number of areas of electron density. Lone pairs count as one area of electron density. Single, double, and triple bonds also only count as one are of electron density. It might be weird to think that a double and triple bond count only as a single area of ele...
Sun Dec 02, 2018 9:57 pm
Forum: Hybridization
Topic: Hybridization of CH3CN
Replies: 2
Views: 157

### Re: Hybridization of CH3CN

The hybridization of the N is sp. This is because there are two region of electron density: the triple bond with the C and the lone pair. The lone pair counts as one area of electron density. The triple bond also only counts as one area of electron density despite there being 6 electrons involved. T...
Sat Nov 24, 2018 11:12 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: 2E. 18
Replies: 1
Views: 83

### Re: 2E. 18

In total, there are 16 valence electrons. After drawing the skeletal structure for N2O with single bonds stemming from the central N to the other N and O, you are left with 12 valence electrons. The way I was taught in high school was to add the remaining electrons to the outer atoms to fill their o...
Sat Nov 24, 2018 11:05 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: T-Shaped
Replies: 5
Views: 248

### Re: T-Shaped

In a T-shaped molecule, there is one bond going straight up, one bond going straight down, and one perpendicular to both of the ones mentioned before that lies on the equatorial plane. The bond angle between the bond going up/down and the bond along the equatorial plane is 90 degrees. The bond angle...
Sat Nov 24, 2018 11:02 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Molecular Shape
Replies: 7
Views: 257

### Re: Molecular Shape

Imagine a molecule with the VSEPR formula, AX4E1. There is a central atom with 4 bonds and one lone pair. The electron geometry for this molecule would be trigonal bipyramidal: one bond going straight up, one bond going straight down, and three on the equatorial plane. However, the formula AX4E1 has...
Thu Nov 15, 2018 12:20 am
Forum: Determining Molecular Shape (VSEPR)
Topic: Bond Angles
Replies: 3
Views: 1211

### Re: Bond Angles

The names that you need to know will most likely include: linear, bent, trigonal planar, tetrahedral, trigonal pyramidal, trigonal bipyramidal, seesaw-shaped, T-shaped, octahedral, square pyramidal, and square planar. The bond angles that you need to know will most likely include: 90, 109.5, 120, 18...
Thu Nov 15, 2018 12:11 am
Forum: Determining Molecular Shape (VSEPR)
Topic: Homework Question 4.1 (6th edition)
Replies: 4
Views: 312

### Re: Homework Question 4.1 (6th edition)

In response to the above post, I don't believe it is possible for the model in part (b) to have two lone pairs. If you think about it, if there were two lone pairs, the electron geometry would be a tetrahedral. Therefore, the two bonds would have a bond angle of slightly less than 109.5 degrees. The...
Thu Nov 15, 2018 12:04 am
Forum: Determining Molecular Shape (VSEPR)
Topic: Molecular Shape 4.91
Replies: 1
Views: 119

### Re: Molecular Shape 4.91

Not quite sure either, but I found this explanation online and it somewhat helps: "Benzyne is an neutral reactive intermediate that can be formed by removing two ortho substituents from a benzene ring leaving the p-orbitals with free electrons. Benzyne is an extremely reactive species because o...
Tue Nov 13, 2018 12:53 am
Forum: Octet Exceptions
Topic: Octet exceptions
Replies: 8
Views: 435

### Re: Octet exceptions

If you look at the principal levels: the 1st principal level only has an s sublevel, the 2nd has an s and a p sublevel, and the 3rd has an s, p, and d sublevel. In order to achieve more than an octet, an atom must have access to the d sublevel in order to go past their octet. This is why atoms in th...
Tue Nov 13, 2018 12:51 am
Forum: Resonance Structures
Topic: Drawing resonance
Replies: 17
Views: 963

### Re: Drawing resonance

My TA said that sometimes the problem won't ask you to draw all of the possible resonance structures even if there are multiple resonance structures. This is because (1) they might not want to give away the fact that there are resonance structures in the specific problem or (2) they aren't asking fo...
Tue Nov 13, 2018 12:45 am
Forum: Formal Charge and Oxidation Numbers
Topic: 3.37 6th edition
Replies: 3
Views: 314

### Re: 3.37 6th edition

For these types of questions, first find out how many valence electrons the unknown element adds to the total amount of electrons in the molecule. Take the total amount of electrons in the molecule and subtract the number of valence electrons of the known elements. For this specific problem, you end...
Thu Nov 01, 2018 2:59 pm
Forum: Trends in The Periodic Table
Topic: Atomic radius across a period
Replies: 3
Views: 172

### Re: Atomic radius across a period

As you travel along a period from left to right, the number of protons and electrons both increase by one. The proton is added to the nucleus and the electron is added to the same sublevel. Since the electrons are added to the same shell, the distance between the charges doesn't increase, but the ch...
Thu Nov 01, 2018 2:38 pm
Forum: Trends in The Periodic Table
Topic: 2.67 Ed 6, electron afinity
Replies: 2
Views: 144

### Re: 2.67 Ed 6, electron afinity

Electron affinity is described as the propensity for an atom or ion to gain an electron. The way that you can tell that C has a higher electron affinity than N is by looking at the 2p sublevels of a C and N atom. C has one electron in two of its three orbitals. N has one electron in each of its orbi...
Thu Nov 01, 2018 2:24 pm
Forum: Trends in The Periodic Table
Topic: Why is ionization energy of oxygen lower than nitrogen?
Replies: 2
Views: 13379

### Re: Why is ionization energy of oxygen lower than nitrogen?

Ionization energy is defined as the amount of energy required to remove an electron from a gaseous atom or ion. In order to answer this question, you need to look at the orbitals of both N and O. If you look at the 2p sublevel for N, there is one electron in each orbital. If you look at the 2p suble...
Sun Oct 28, 2018 11:35 pm
Forum: Properties of Electrons
Topic: 1E.7 for the HW
Replies: 2
Views: 362

### Re: 1E.7 for the HW

You can tell if an atom is in an excited by looking at the end of the electron configuration. If you find an electron that is in a higher energy level than where it typically should be, it is most likely in an excited state. Take neon (Ne) for example: The normal electron configuration would look li...
Sun Oct 28, 2018 11:29 pm
Forum: Einstein Equation
Topic: Calculating number of photons emitted
Replies: 3
Views: 646

### Re: Calculating number of photons emitted

The Einstein equation only accounts for the energy of a single photon of light, so if you had no other information, you wouldn't really be able to answer the equation. All you need to find the total number of photons is the total energy of the light source. You take the total energy and divide that ...
Sun Oct 28, 2018 11:25 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Calculating Velocity
Replies: 4
Views: 211

### Re: Calculating Velocity

Don't worry if you get a super small or large value for your delta v or delta x. As long as you plugged in the correct numbers then you should get the correct value in the end. The size of the value only determines how big the range of uncertainty is. So, if you end up getting a super small value fo...
Sat Oct 20, 2018 2:04 am
Forum: Properties of Electrons
Topic: Light vs Electron
Replies: 1
Views: 58

### Re: Light vs Electron

Electrons DO exhibit wave particle duality. It was proved through an experiment by Davisson and Germer. It's kind of hard to explain so I'll just give you a link to a video I watched to better understand it.

Sat Oct 20, 2018 1:48 am
Forum: Properties of Light
Topic: color of light
Replies: 11
Views: 450

### Re: color of light

If you look at a typical spectrum with the range of visible light, you will find that many spectrums use wavelength to differentiate between colors. However, since wavelength and frequency are inversely related, you can say that frequency also determines the color of light.
Sat Oct 20, 2018 1:44 am
Forum: Properties of Light
Topic: 6th Edition: Problem 1.9
Replies: 3
Views: 158

### Re: 6th Edition: Problem 1.9

I used the chart that was found a few pages before the questions to identify which event correlated to which wavelength. I'm not sure if we have to memorize those sections though.
Sat Oct 20, 2018 1:40 am
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: Quantum Number (ml)
Replies: 1
Views: 89

### Quantum Number (ml)

I know that the ml quantum number is the magnetic quantum number, and that it can be px, py, or pz, and it represents the orientations of different orbitals of a subshell. But what do those things mean? Thanks in advance.
Sat Oct 20, 2018 1:34 am
Forum: Balancing Chemical Reactions
Topic: Chem Test Balancing Equations
Replies: 2
Views: 401

### Re: Chem Test Balancing Equations

In order to balance this equation, you first need to recognize the reactants and products. The reactants are C7H6O3 and O2, and the products are CO2 and H2O. So first, write the equation as follows: C7H6O3 + O2 ---> CO2 + H2O Now comes the part where you balance this equation. Since both the C and t...
Sat Oct 20, 2018 1:24 am
Forum: Empirical & Molecular Formulas
Topic: Polyatomic Ions and Naming
Replies: 3
Views: 366

### Re: Polyatomic Ions and Naming

This chart contains most of the common polyatomic ions that I've needed to memorize in high school! I'm not sure how much more we'll need to know for this class though. https://www.google.com/search?q=list+of+common+polyatomic+ions&safe=strict&source=lnms&tbm=isch&sa=X&ved=0ahUKE...
Sat Oct 20, 2018 1:16 am
Forum: Molarity, Solutions, Dilutions
Topic: Online Module Problem Dilutions
Replies: 2
Views: 531

### Re: Online Module Problem Dilutions

First, you need to find the molarity of the original KMnO4 solution. In order to find molarity, you need moles of solute and liters of solution. Use the molar mass of KMnO4 to convert grams to moles, and divide that number by .15000 L. You should end up getting .211 M KMnO4. Second, you need to use ...
Fri Oct 05, 2018 3:25 am
Forum: Significant Figures
Topic: Sig figs for Avogadro's number
Replies: 3
Views: 107

### Re: Sig figs for Avogadro's number

In high school chemistry, I typically used 6.02*10^23 as Avogadro's number, but in lecture I remember Lavelle saying that we should use 6.022*10^23. I guess we can use that extra sig fig now that we are grown college students.
Fri Oct 05, 2018 3:22 am
Forum: Significant Figures
Topic: Sig Figs
Replies: 3
Views: 232

### Re: Sig Figs

In your first multiplication calculation (1*12.011), you will end up with 5 significant figures. In your second multiplication calculation, (2*1.008), you will end up with 4 significant figures. However, in your last calculation, you adhere to the addition/subtraction rule for significant figures.Wh...
Fri Oct 05, 2018 3:15 am
Forum: Significant Figures
Topic: Using Sig Figs Throughout a Problem?
Replies: 6
Views: 217

### Re: Using Sig Figs Throughout a Problem?

Whenever you are doing a multi-step problem, it is always good to round at the very end. If you round after each step, your answer may come out to be very different and inaccurate. Try your best to keep all of your answers in your calculator so you can go back and use the exact answers in all subseq...