## Search found 101 matches

Sat Mar 13, 2021 8:43 am
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: Catalysts and intermediates
Replies: 10
Views: 49

### Re: Catalysts and intermediates

Intermediates will be produced first and then consumed, so they will appear first on the right and then on the left of the reaction. On the other hand, catalysts will be consumed first and then produced, so they will appear first on the left and then on the right. Hope this helps!
Sat Mar 13, 2021 8:34 am
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Help with Thermochem UA problem
Replies: 4
Views: 30

### Re: Help with Thermochem UA problem

Hi! Would you mind showing your work for how you got work? I think your ∆n might be off (∆n should be 0.5, but it seems like you might have plugged in 1?). Yeah, I plugged in 1 for n because it is 1 mol of CO2. Why would n be 0.5? Delta n would be .5 because you’re looking for the change in moles o...
Fri Mar 12, 2021 9:36 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: When to use Platinum
Replies: 14
Views: 57

### Re: When to use Platinum

You would use platinum when there is no metal conductor in the half reaction. Another thing to note is that mercury does not require a platinum even though it is in its liquid state at room temperature, since it is still a conducting metal. Hope this helps!
Fri Mar 12, 2021 9:23 pm
Forum: Second Order Reactions
Topic: Determining slow step
Replies: 19
Views: 91

### Re: Determining slow step

Intermediates which are produced first and then consumed will not be included in the rate law even if they are in the slow step. To answer your second question, the slow step will have the highest activation energy and its reactants will be in the final rate law. Hope this helps!
Wed Mar 10, 2021 1:59 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Strongest reducing metal
Replies: 8
Views: 32

### Re: Strongest reducing metal

The strongest reducing metal will be the one that is most likely to be oxidized, as it must cause the reduction of another molecule. Based on this, the metal with the most negative reduction potential will be the strongest reducing metal since it itself is the least likely to be reduced. Hope this h...
Sat Mar 06, 2021 6:59 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 6M.5 Pt on both sides?
Replies: 3
Views: 16

### Re: 6M.5 Pt on both sides?

In a similar post, Dr. Lavelle was mentioning that mercury is the only conducting metal that is a liquid at room temperature, so I think this is why you don't need to add a Pt. Hope this helps!
Sat Mar 06, 2021 6:48 pm
Forum: Zero Order Reactions
Topic: Half life
Replies: 20
Views: 149

### Re: Half life

To calculate half life of a zero order reaction, you would use the equation: t1/2 = [A0] / 2k where A0 is the initial concentration and k is the rate constant.
Sat Mar 06, 2021 6:09 pm
Forum: Balancing Redox Reactions
Topic: Oxidizing Agent
Replies: 33
Views: 96

### Re: Oxidizing Agent

The best oxidizing agent will have the highest reduction potential, while the weakest oxidizing agent will have the lowest reduction potential since it is not as likely to be reduced and cause the oxidation of another molecule.
Sat Mar 06, 2021 5:57 pm
Forum: Balancing Redox Reactions
Topic: 6k.5
Replies: 2
Views: 8

### Re: 6k.5

There’s actually an error in this question - in the revised Solution Manual Errors document, it mentions that this problem should actually be Cl2(g) —> HClO(aq) + Cl- (aq), so you shouldn't get Cl2 on both sides. Hope this helps!
Sat Mar 06, 2021 5:50 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: E naught
Replies: 8
Views: 53

### Re: E naught

E naught will be 0 in concentration cells. This is because the reactions occurring at the anode and cathode are equivalent, so if their concentrations are the same (which must be true at standard conditions), their potentials will cancel each other out and equal 0. Hope this helps!
Sat Feb 27, 2021 10:52 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: Nernst K vs Q
Replies: 14
Views: 224

### Re: Nernst K vs Q

If you’re using Q with the Nernst equation, it means that your reaction is not at equilibrium, but by using K (the equilibrium constant), it means your reaction is at equilibrium.
Sat Feb 27, 2021 10:42 pm
Forum: Balancing Redox Reactions
Topic: Determining which molecule is the oxidizing agent
Replies: 49
Views: 148

### Re: Determining which molecule is the oxidizing agent

The oxidizing agent basically causes another molecule to be oxidized, which means it itself is reduced. For example, in order for another molecule to be oxidized and lose a negative charge, the oxidizing agent must be the one to accept that negative charge, so it becomes reduced.
Sat Feb 27, 2021 10:36 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Salt Bridges
Replies: 41
Views: 134

### Re: Salt Bridges

The salt bridge prevents a buildup of charge on one side of the galvanic cell and essentially maintains a neutral environment to allow the reaction to continue.
Sat Feb 27, 2021 10:10 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: Sapling #15
Replies: 5
Views: 60

### Re: Sapling #15

Your work looks right to me! Since the other molecules are solids, you don’t include them in your value for Q, so you have all the information you need to solve the problem.
Sat Feb 27, 2021 7:20 pm
Forum: Student Social/Study Group
Topic: Final
Replies: 63
Views: 248

### Re: Final

I definitely agree that doing the textbook problems again is a great way to study, especially for the math based problems; for the conceptual problems, I also usually read the textbook skimming for any concepts that I hadn’t come across anywhere else. If there’s a topic I really can’t seem to unders...
Sat Feb 20, 2021 7:42 pm
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: Degeneracy and Entropy
Replies: 6
Views: 47

### Re: Degeneracy and Entropy

Since degeneracy is the amount of possible ways that the molecule can exist at a certain energy state, the randomness increases with degeneracy because those molecules can exist in any one of the various possible states within a system. On the other hand, if a molecule can only exist in one orientat...
Sat Feb 20, 2021 7:31 pm
Forum: Calculating Work of Expansion
Topic: unit of w
Replies: 17
Views: 89

### Re: unit of w

Whenever you use this work equation, you’ll get units of liters times atmospheres. However, if the answer is in Joules or you need to combine this equation with another value, like with heat to find delta U, you’ll need to convert the units to Joules. You can do this by multiplying your answer for w...
Sat Feb 20, 2021 7:25 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Delta H and Delta S both positive
Replies: 31
Views: 128

### Re: Delta H and Delta S both positive

If both delta H and delta S are positive, then the spontaneity will depend on temperature. Looking at the deltaG = deltaH - TdeltaS equation, having a positive enthalpy and entropy means that if temperature is large enough, it will make delta G negative, making the reaction spontaneous. However, if ...
Sat Feb 20, 2021 7:03 pm
Forum: Van't Hoff Equation
Topic: Application of Van't Hoff Equation
Replies: 11
Views: 62

### Re: Application of Van't Hoff Equation

I think it's most helpful for finding the equilibrium constant for a reaction at a different temperature, but you can basically use it to solve for whatever term you want that’s in the equation, as long as you know all the other variables. For example, if you already know both equilibrium constants,...
Sat Feb 20, 2021 6:55 pm
Forum: Student Social/Study Group
Topic: Post Midterm 2 De-stressing
Replies: 92
Views: 543

### Re: Post Midterm 2 De-stressing

I’m planning on de-stressing by making myself some hot chocolate and catching up on all the tv shows I haven’t had time to watch!
Sat Feb 13, 2021 10:03 pm
Forum: Student Social/Study Group
Topic: Sapling Week 5-6 Homework Question #7
Replies: 4
Views: 34

### Re: Sapling Week 5-6 Homework Question #7

Hi Mrudula! So the way I solved this problem was basically by looking at the units of each answer and working backwards. The first thing you have to do to determine enthalpy is find out how many kilojoules of energy are being supplied to the system. Using 550 W (which is equal to 550 J/s) you can co...
Sat Feb 13, 2021 9:50 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Sapling Question #16
Replies: 8
Views: 67

### Re: Sapling Question #16

You shouldn’t need to use temperature in your calculations, since it’s basically just there to let you know that the reaction is occurring at standard conditions, which is consistent with the values from the chart given to you. To solve the problem, you can just add up the sum of the deltaG of forma...
Sat Feb 13, 2021 9:42 pm
Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
Topic: Textbook Problem 4H.1
Replies: 4
Views: 22

### Re: Textbook Problem 4H.1

Some general guidelines for how to determine what has higher entropy is by looking at the states of the molecules. So liquids will have higher entropy than solids, and gases - which are very disordered - will have higher entropy than liquids. You can also look at degeneracy, so a compound with a lot...
Sat Feb 13, 2021 9:29 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Sapling question 4
Replies: 9
Views: 32

### Re: Sapling question 4

The way I solved this problem was by using deltaS = nRln(V2/V1). To find the values for V2 and V1, I basically wrote V2 in terms of V1, so V2 = (V1*⅙). When you plug these expressions into the ln function, the V1 variables cancel out and leave you with a number. Then, by using n=1, you should be abl...
Sat Feb 13, 2021 9:17 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Is it necessary to memorize Cp,m or Cv,m values? [ENDORSED]
Replies: 26
Views: 123

### Re: Is it necessary to memorize Cp,m or Cv,m values?[ENDORSED]

Like others have said, I’m pretty sure all the values/formulas we’ll need are on the equation sheet, so you shouldn't have to worry too much about memorization. I would also add that the expressions with coefficients in front of R on the equation sheet are for monatomic ideal gases, so we might have...
Sat Feb 06, 2021 8:08 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: How to know the sign for work
Replies: 26
Views: 156

### Re: How to know the sign for work

Work is negative when the system itself is doing work (because the system requires energy and therefore loses it); conversely, work is positive when the surroundings are doing work on the system (because energy is being applied to the system).
Sat Feb 06, 2021 7:41 pm
Forum: Phase Changes & Related Calculations
Topic: sapling wk 3/4 #13
Replies: 3
Views: 41

### Re: sapling wk 3/4 #13

Like Steph said, the equation you would use to solve this problem would be w = -nRTln(Vf/Vi) where Vf is the final volume and Vi is the initial volume. You are given all the values in this equation except for moles (n). To find this, you would use the equation PV = nRT and plug in the values given t...
Sat Feb 06, 2021 7:29 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Solving for Cubic Equations
Replies: 13
Views: 44

### Re: Solving for Cubic Equations

Cubic equations are not the only instance where you can make X negligible. You can use this process anytime the equilibrium constant is less than 10^-4. You can also use the 5% rule by making sure that your final concentration is less than 5% than the initial concentration. If it is, then you can co...
Sat Feb 06, 2021 7:14 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: sapling
Replies: 3
Views: 22

### Re: sapling

The right side of your equation looks correct but the left side should actually have two terms: one to represent the ice melting and the other to represent the melted ice heating up. The melting of ice can be represented as moles times enthalpy of fusion. (we use moles instead of mass because enthal...
Sat Feb 06, 2021 7:04 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: isothermal reactions
Replies: 9
Views: 58

### Re: isothermal reactions

An isothermal reaction means that temperature remains constant throughout the reaction; however, there is still a transfer of heat, energy, etc. Hope this helps!
Sat Jan 30, 2021 11:31 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 4E.5 C-C Bond Enthalpy
Replies: 6
Views: 50

### Re: 4E.5 C-C Bond Enthalpy

Like others have said, the reason they calculate 6 bonds using the same bond enthalpy is because of resonance, as the electrons are shared among the carbons. The answers have also disregarded the 6 C-H bonds because they are not being broken; however, I would also add that you can still get the same...
Sat Jan 30, 2021 11:23 am
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: percent protonation/deprotonation
Replies: 14
Views: 112

### Re: percent protonation/deprotonation

% deprotonation refers to when acids donate an H+ ion to water, as they are losing that proton. % protonation is the opposite and refers to bases when they are gaining a proton. You can calculate these two values by dividing the H3O+ or OH- concentrations at equilibrium by the initial concentration ...
Sat Jan 30, 2021 11:10 am
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: How Does Temperature Affect the Equilibrium Constant?
Replies: 25
Views: 135

### Re: How Does Temperature Affect the Equilibrium Constant?

If the reaction is exothermic and heat is being released, this means that the reaction will shift towards the reactants to account for this change. Since reactants are in the denominator of the K expression, increasing temperature will cause K to decrease. The opposite is true for exothermic reactio...
Sat Jan 30, 2021 11:06 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: steam causing burns
Replies: 40
Views: 216

### Re: steam causing burns

The reason steam causes higher burns is because it undergoes condensation when it hits our skin. This is represented by the second horizontal line in the phase change diagram, which covers a significant amount of heat released, specifically 46kJ. Since condensation releases this much energy, it caus...
Sat Jan 30, 2021 10:57 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Diatomic molecules standard enthalpy of formation
Replies: 22
Views: 115

### Re: Diatomic molecules standard enthalpy of formation

Yes, most of the diatomic molecules will be equal to 0 if they are in the gas phase. This is because they are in their standard state. However, I think a diatomic bromine molecule would be zero when it is in its liquid state whereas the diatomic iodine molecule has to be in a solid state, as these m...
Sat Jan 23, 2021 3:58 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Weak acids/ Weak Bases
Replies: 3
Views: 50

### Re: Weak acids/ Weak Bases

In addition to what Hila already mentioned, there are also some common signals that you can recognize to determine whether the compound is an acid or base. For example, if a compound ends in COOH, it is most likely a weak acid, and if a compound has a Nitrogen with a lone pair, it is most likely a w...
Sat Jan 23, 2021 3:49 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: X Approximations
Replies: 23
Views: 113

### Re: X Approximations

If the value of the equilibrium constant is less than 10^-4 you can assume that it will be small enough to approximate. You can also use the 5% rule, so if your x value is less than 5% of the initial concentration, you should be able to approximate. Hope this helps!
Fri Jan 22, 2021 10:54 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: H2O and K
Replies: 11
Views: 59

### Re: H2O and K

Yes, we only exclude H2O in equilibrium constants if it is in solid or liquid form since there is so much present before and after the reaction that the change in H2O is negligible. This applies to substances other than water as well, so any substance in the form of a liquid or solid should not be i...
Fri Jan 22, 2021 10:37 pm
Forum: General Science Questions
Topic: Lecture 6 question
Replies: 2
Views: 8

### Re: Lecture 6 question

I think the Kb value should actually be 1.8x10^-5 with a negative exponent instead of a positive one. If you use this number you should get the right answer, so I don’t think you’re doing anything wrong with your calculator. Hope this helps!
Fri Jan 22, 2021 10:31 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Ordering from weakest to strongest Acd/base
Replies: 11
Views: 90

### Re: Ordering from weakest to strongest Acd/base

A general trend is that a stronger acid has a lower pKa and a higher Ka. This is because strong acids dissociate completely, so a strong acid should have a high dissociation constant. The same trend can be said for bases, so stronger bases would have a lower pkb and higher Kb.
Fri Jan 15, 2021 3:29 pm
Forum: Student Social/Study Group
Replies: 5
Views: 68

### Re: Answering Questions with Responses

Yes, like others have said, you can still get credit for answering questions that already have multiple answers. I actually find it helpful since everyone learns differently, so certain explanations might make more sense than others to different people. Hope this helps!
Fri Jan 15, 2021 3:14 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Temperature
Replies: 45
Views: 179

### Re: Temperature

The question will likely tell you what delta H is and you can use this to determine whether the reaction is endothermic or exothermic and how it will react to changes in temperature. If the question doesn’t give you a value for delta H, a general trend is that when you’re forming bonds, you release ...
Wed Jan 13, 2021 9:33 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Several Questions
Replies: 3
Views: 44

### Re: Several Questions

Yes! For your first question you're right, [H+] and [H3O+] mean the same thing. Your second question is right as well, you only include H20 in the equilibrium equation when it is in gas form, not when it is in solid or liquid form. Hope this helps!
Tue Jan 12, 2021 1:48 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Reaction Quotient
Replies: 10
Views: 495

### Re: Reaction Quotient

K is the ratio of products to reactants only when the reaction is at equilibrium, while Q is the ratio of products to reactants at a random point in time. By comparing Q and K, we can find out what direction a given reaction (Q) shifts towards in order to reach equilibrium (K). For example, if Q is ...
Mon Jan 11, 2021 5:11 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Textbook Problem 5.J.5 part B
Replies: 5
Views: 48

### Re: Textbook Problem 5.J.5 part B

When looking at a reaction’s response to an increase in pressure or decrease in volume, you’ll want to look at the moles of the equation. The reaction will shift towards the side with less moles, which in this case would be towards the left. This is because there are two moles of gas on the right an...
Thu Jan 07, 2021 6:05 pm
Forum: Administrative Questions and Class Announcements
Replies: 7
Views: 89

Like others have said, only some UAs give out an answer key for their worksheets, but if you can’t find them anywhere you could always ask about some specific questions on Chem Community for some help and easy points! Hope this helps!
Thu Jan 07, 2021 5:57 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: concentration or partial pressure
Replies: 7
Views: 39

### Re: concentration or partial pressure

Like others have said, you wouldn’t have to convert any values in this case since they are already given in the units of concentration (mol/L). If the units they gave you were in terms of atm or bars, this would indicate partial pressure and you would use PV = nRT in order to convert these values. H...
Thu Jan 07, 2021 5:50 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Direction of a non-equilibrium reaction
Replies: 11
Views: 101

### Re: Direction of a non-equilibrium reaction

If Q is less than K, this means that the value of the denominator (reactants) is too big and the numerator (products) is too small. In other words, there are too many reactants in the reaction compared to how much there needs to be in order for the reaction to be at equilibrium. Therefore, in order ...
Thu Jan 07, 2021 5:39 pm
Forum: Ideal Gases
Topic: PV=nRT
Replies: 74
Views: 549

### Re: PV=nRT

when do you know you are allowed to use this formula? You would be using this formula if you want to convert between partial pressures and concentrations. For example, if the problem asks you to solve for Kc but you are only given partial pressures, you can plug these values into the equation to ge...
Thu Jan 07, 2021 5:27 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Partial pressure vs concentration
Replies: 6
Views: 65

### Re: Partial pressure vs concentration

Like others have said, both Kp and Kc can be used when calculating an equilibrium constant involving a reaction of gases. However, I think Kp is typically used for calculating the equilibrium constant for these reactions with only gases as opposed to aqueous solutions. Hope this helps a little!
Fri Dec 11, 2020 12:54 pm
Forum: Student Social/Study Group
Topic: Anxiety
Replies: 75
Views: 499

### Re: Anxiety

Whenever I'm feeling really anxious about something, I like to take a break by going on a walk and just listening to music. It really helps me take my mind off of what is stressing me out and just overall helps me feel more refreshed.
Fri Dec 11, 2020 11:40 am
Forum: Properties of Electrons
Topic: How can an electron be excited?
Replies: 4
Views: 49

### Re: How can an electron be excited?

In the photoelectric effect, an electron will be released if the incoming photon is higher in energy than the threshold energy. If you’re talking about atomic spectra, then an incoming photon must have energy that exactly matches the difference in energy between two energy levels in order for the el...
Thu Dec 10, 2020 2:29 pm
Forum: Electronegativity
Topic: solubility
Replies: 2
Views: 25

### Re: solubility

The more ionic character a molecule has, the more soluble it will be. I’m not too sure why that is though so maybe someone else can elaborate on this!
Thu Dec 10, 2020 2:25 pm
Forum: Naming
Topic: Textbook Question 9C.1
Replies: 1
Views: 22

### Re: Textbook Question 9C.1

I think it helps to understand if you do an example; for instance, cisplatin: [Pt(Cl)2(NH3)2]. Since the charge of the compound is neutral (or 0), you know that when you add all the charges of the molecules inside the brackets, they should also equal 0. We know that the charge of Cl is -1, and since...
Thu Dec 10, 2020 10:44 am
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Early Quantum Question (from Wilberth's Q&A)
Replies: 4
Views: 92

### Re: Early Quantum Question (from Wilberth's Q&A)

What confused me at first was the vocabulary of “first excited state and second excited state” so maybe this is your problem too? The first excited state would be n=2, (not 1 since n=1 is the ground state), and the second excited state would be n=3. So you would use the equation -deltaE = -hR/nf^2 -...
Wed Dec 02, 2020 1:04 pm
Forum: Administrative Questions and Class Announcements
Topic: Chem 14A Final
Replies: 21
Views: 230

### Re: Chem 14A Final

In the Week 10 review session pdf it said the final would be weighted based on how much time we spent on each unit, so I don’t think there will be a super large focus only on the last units. I think he said that the quantum unit was the longest so we should expect more questions from that section. H...
Wed Dec 02, 2020 10:48 am
Forum: Lewis Structures
Topic: 2E.25 Part A
Replies: 2
Views: 35

### Re: 2E.25 Part A

I don’t think it matters where you place them, since either way, you’ll end up with a polar molecule. Although it doesn’t look like it in the Lewis structure, none of the atoms in a tetrahedral shape are located directly across from each other (every bond angle is 109 not 180), so no matter where yo...
Wed Dec 02, 2020 9:38 am
Forum: Naming
Topic: Ligand Names Table on Final
Replies: 10
Views: 106

### Re: Ligand Names Table on Final

In the pdf about Week 10 final review sessions, it says the only resources we can have on the final are a calculator, the periodic table, and the constants/equations sheet, so I think we’ll have to memorize the ligand names since we won’t have it for the test. Hope this helps!
Wed Dec 02, 2020 9:32 am
Forum: Naming
Topic: Ferrate
Replies: 7
Views: 111

### Re: Ferrate

Yes, you would use “ferrate” instead of “ironate” in coordination compounds when the overall charge of the compound is negative. However, if there is no charge or the compound is positive, you would just use “iron.” Hope this helps!
Tue Dec 01, 2020 1:48 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Textbook Problem 2E #13
Replies: 3
Views: 75

### Re: Textbook Problem 2E #13

In the answer key that explains how to solve each problem it says the answer is AX2E3, so maybe there’s a typo in your book? The middle iodine would be the central atom A, leaving the two other iodine atoms to be X2. There are also three lone pairs on the central iodine, so E would be three. Hope th...
Tue Nov 24, 2020 8:49 pm
Forum: Hybridization
Topic: Electron Configuration
Replies: 7
Views: 434

### Re: Electron Configuration

Is there a rule to these exceptions or do we just need to memorize them? Yes, only the atoms that are one electron away from the 3d5 or 3d10 orbitals would take away an electron from the 4s subshell to become more stable, which is why chromium and copper are exceptions. I’m pretty sure this same tr...
Tue Nov 24, 2020 8:42 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Sapling Question #1
Replies: 7
Views: 71

### Re: Sapling Question #1

Maddie Turk Disc 2C wrote:So just to clarify, AX4 will always signify a tetrahedral shape, alone with AX3E always being trigonal pyramidal, correct? almost as if they are code for that shape?

You're right! Molecules with the same VSEPR notation will always have the same shape no matter what the atoms are.
Tue Nov 24, 2020 6:29 pm
Forum: Administrative Questions and Class Announcements
Replies: 4
Views: 45

I'm pretty sure 465/500 points would be an A, if 93% is the cutoff. For an A- I think 450/500 would get you a 90% and 480/500 would get you an A+ if 96% is the cutoff. This could change if the cutoffs are different or a curve happens, so just keep that in mind! Hope this helps!
Tue Nov 24, 2020 12:00 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Polar Bonds vs. Polar Molecules
Replies: 4
Views: 52

### Re: Polar Bonds vs. Polar Molecules

A polar bond refers to the difference in electronegativity between two elements, while a polar molecule refers to whether a molecule overall carries a charge. For example, in CH3Cl, there is a polar bond between the C and Cl due to the large difference in electronegativity; this dipole causes the en...
Tue Nov 24, 2020 11:53 am
Forum: Determining Molecular Shape (VSEPR)
Topic: Sapling Question #1
Replies: 7
Views: 71

### Re: Sapling Question #1

Although both the tetrahedral structure and SO32- have five areas of electron density, SO32- has a lone pair where a tetrahedral has its fifth element. Although lone pairs can influence shape, only atom positions determine the actual naming of the molecule’s shape, so SO32- creates a trigonal pyrami...
Mon Nov 23, 2020 2:58 pm
Forum: Administrative Questions and Class Announcements
Topic: Discussion
Replies: 2
Views: 35

### Re: Discussion

Like Ria said, there shouldn’t be any discussions after Wednesday this week. I don't think those discussions will have to make up anything since the Thursday and Friday discussions are actually ahead right now because we started this fall quarter on a Thursday. After this short week though, we shoul...
Fri Nov 20, 2020 1:59 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: VSEPR notation
Replies: 9
Views: 37

### Re: VSEPR notation

Yes! To build off of what others have said, molecules with the same VSEPR notation will have the same shape. For example, all molecules with the AX3E notation will have a trigonal pyramidal structure. Hope this helps!
Fri Nov 20, 2020 1:31 pm
Forum: Ionic & Covalent Bonds
Topic: Polar Bonds
Replies: 19
Views: 154

### Re: Polar Bonds

You can determine this by looking at the differences in electronegativity between the atoms; atoms with a large difference in electronegativity will create a dipole and cause the molecule to be polar. For example, in an HF molecule, the large difference in electronegativity would cause electrons to ...
Fri Nov 20, 2020 1:25 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Electron Configuration for Silver?
Replies: 8
Views: 86

### Re: Electron Configuration for Silver?

Oh okay. For V; Vanadium does the 4s2 give electrons to the 3d3 to make a half full 3d orbital then? Vanadium’s electron configuration would actually still be [Ar] 3d3 4s2. Only the atoms that are one electron away from 3d5 or 3d10 would take away an electron from 4s, so only the groups of chromium...
Fri Nov 20, 2020 9:12 am
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: Determining an orbital
Replies: 10
Views: 108

### Re: Determining an orbital

Like others have said, the quantum number l determines the shape of the orbital, which is how s, p, or d are determined. The ml quantum number tells you the orientation that these shapes are in; for example, in the p orbital, there would be 3 orientations where the two lobes lie on either the x axis...
Thu Nov 19, 2020 7:51 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Electron configuration exceptions
Replies: 7
Views: 104

### Re: Electron configuration exceptions

I’m pretty sure you’re right about Nickel having the electron configuration [Ar]3d8 4s2, so I’m not too sure why it would be 3d10. Based on the electron configuration of [Ar]3d8 4s2 though, Ni 2+ would be [Ar] 3d8, since you would be removing 2 electrons from the outermost orbital 4s. Also the elect...
Thu Nov 12, 2020 4:27 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: textbook problem 2A #5
Replies: 2
Views: 35

### Re: textbook problem 2A #5

Copper’s ground state electron configuration is actually an exception; instead of leaving only one unpaired electron in the 3d subshell, it fills up the entire subshell and leaves one 4s orbital unfilled. So the normal electron configuration for copper is actually [Ar]3d^10 4s^1. Since the problem a...
Thu Nov 12, 2020 3:33 pm
Forum: Polarisability of Anions, The Polarizing Power of Cations
Topic: Polarizability
Replies: 6
Views: 45

### Re: Polarizability

Between two anions in the same row, would an anion with a -2 charge be more or less polarizable than an anion with a -1 charge? I think the anion with the -2 charge would be more polarizable because it would be the larger atom. For example, O^2- would be more polarizable than F^- because atomic rad...
Thu Nov 12, 2020 3:24 pm
Forum: Octet Exceptions
Topic: How to tell
Replies: 6
Views: 108

### Re: How to tell

An element can have an expanded octet if its energy level has access to a d orbital. For example, the elements in the 3rd period have access to the 3d subshell, so elements like Phosphorus and Sulfur can have more than the traditional octet. However, elements in the second period like Nitrogen or Ox...
Thu Nov 12, 2020 8:41 am
Forum: Polarisability of Anions, The Polarizing Power of Cations
Topic: Polarizability
Replies: 9
Views: 57

### Re: Polarizability

Polarizability is how easily an anion’s electrons can be pulled towards a cation. The more polarizable an anion is, the more covalent character the ionic bond has. The main factor that determines polarizability is size; larger atoms like iodine are more polarizable because their electrons are farthe...
Wed Nov 11, 2020 11:01 am
Forum: Trends in The Periodic Table
Topic: Textbook Problem Ionization Energy Oxygen
Replies: 3
Views: 48

### Re: Textbook Problem Ionization Energy Oxygen

It has a lower ionization energy because of electron repulsion. In an Oxygen atom, the 2p subshell has four electrons: 2px^2, 2py^1, 2pz^1. Since the 2px orbital has 2 electrons, these electrons repel each other, making it easier for one of them to be removed. This is why oxygen and the other elemen...
Fri Nov 06, 2020 9:20 pm
Forum: Lewis Structures
Topic: lewis structure
Replies: 10
Views: 112

### Re: lewis structure

Like Justin said, the formula for formal charge equals valence electrons - (shared electrons/2 + lone pairs). To symbolize the +1 formal charge of Nitrogen on the actual Sapling problem, you have to put the green positive symbol under “More” onto the Nitrogen atom. Hope this helps!
Fri Nov 06, 2020 7:54 pm
Forum: Student Social/Study Group
Topic: Blind sided by Midterm 1 memorization questions, How to study for memorization questions
Replies: 11
Views: 118

### Re: Blind sided by Midterm 1 memorization questions, How to study for memorization questions

You’re not alone - the conceptual questions really threw me off on the midterm too! Like everyone else said, taking notes is a solid way to study and absorb the material. I think something else that might be helpful is to look up explanations of certain concepts on other websites and sources - they ...
Fri Nov 06, 2020 6:42 pm
Forum: Formal Charge and Oxidation Numbers
Topic: Formal Charge vs. More/Less Valence Electrons
Replies: 3
Views: 29

### Re: Formal Charge vs. More/Less Valence Electrons

When I went to one of the workshops, the UA said to prioritize making a full octet over making all formal charges zero. So I don’t think you should create an expanded octet just to get all formal charges equal to zero. I’m still not 100% confident on this concept though, so hopefully someone else ca...
Fri Nov 06, 2020 2:19 pm
Forum: Administrative Questions and Class Announcements
Topic: Midterm Curve?
Replies: 12
Views: 142

### Re: Midterm Curve?

Yeah, I don’t think the class is curved. The class is out of a total of 500 points, so I think your percentage is based purely on how many points you get out of 500.
Fri Nov 06, 2020 9:53 am
Forum: Ionic & Covalent Bonds
Replies: 38
Views: 267

The atomic radius decreases as you go across a period because more protons are being added to the nucleus as you go across. With every proton added, the attraction between the electrons and protons grows, bringing the electrons closer and closer to the nucleus. Therefore, the atomic radius gets smal...
Thu Oct 29, 2020 8:11 pm
Forum: Student Social/Study Group
Topic: Test Scored for Midterm 1
Replies: 49
Views: 401

### Re: Test Scored for Midterm 1

I heard that they'll be released when the last discussion section is done taking it. I didn't hear it from Dr. Lavelle though so I could be wrong! My guess would be that you could find the score on CCLE under the midterm section.
Thu Oct 29, 2020 7:22 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: h vs. ħ?
Replies: 11
Views: 187

### Re: h vs. ħ?

I think the h-bar symbol represents h/2pi, so (½)h-bar would be equal to h/4pi. Since both equations are basically equivalent to each other, I’m pretty sure you can use whichever form you’re most comfortable with. Hope this helps!
Thu Oct 29, 2020 6:15 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Exercise 1A #15
Replies: 2
Views: 22

### Re: Exercise 1A #15

The first thing I did was figure out the change in energy using the equation deltaE = - (hc)/wavelength. (The value is negative since it is an emission spectrum, so it releases energy). We know that delta E = E final - E initial, or (-hR/nf^2) - (-hR/ni^2), where nf stands for the final energy level...
Thu Oct 29, 2020 6:01 pm
Forum: Administrative Questions and Class Announcements
Topic: Chem 14A Website
Replies: 5
Views: 87

### Re: Chem 14A Website

I'm pretty sure the password is LL14A20. Hope this works!
Thu Oct 29, 2020 10:39 am
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Atomic Spectra
Replies: 4
Views: 45

### Re: Atomic Spectra

You’re right in that the electron does absorb energy from the photon, which causes the electron to go up an energy level. However, for the electron to come back down, it has to release the energy of the photon that it absorbed (which is recorded to get the emission spectrum). Hope this helps!
Wed Oct 21, 2020 3:13 pm
Forum: DeBroglie Equation
Topic: De Broglie Wavelength vs Wavelength based on EM Spectrum
Replies: 5
Views: 84

### Re: De Broglie Wavelength vs Wavelength based on EM Spectrum

I think this is because light on the electromagnetic spectrum does not have a mass, so you wouldn’t be able to apply the de broglie equation to it. However, for objects with a small mass, like electrons, you can calculate the de broglie wavelength. Hope this helps!
Wed Oct 21, 2020 9:33 am
Forum: Photoelectric Effect
Topic: Sapling Homework Question 10
Replies: 3
Views: 44

### Re: Sapling Homework Question 10

It would actually be more helpful to use the de Broglie equation (wavelength = h/mv)! Since you already have the values for h and v (velocity), the only thing you need to solve for is the mass of an individual diatomic chlorine molecule. Hope this helps!
Wed Oct 21, 2020 9:27 am
Forum: DeBroglie Equation
Topic: Sapling Question #10
Replies: 8
Views: 112

### Re: Sapling Question #10

The equation I used for this was the De Broglie equation: wavelength = h/mv. We already know the values of h and v (velocity) so the only thing we need to calculate is the mass. Since it’s a diatomic fluorine molecule, we can multiply the molar mass of fluorine by two to get the molar mass of the mo...
Mon Oct 19, 2020 1:47 pm
Forum: Properties of Electrons
Topic: Photoelectric Experiment
Replies: 2
Views: 17

### Re: Photoelectric Experiment

The photoelectric experiment was originally designed to test how much energy it would take to eject an electron from a piece of metal. However, they ended up finding out that in the experiment, light acted as a photon rather than a wave.
Mon Oct 19, 2020 1:44 pm
Forum: Properties of Light
Topic: Photoelectric Experiment
Replies: 3
Views: 12

### Re: Photoelectric Experiment

The surprising factor of the photoelectric experiment was that light did not act like a wave, but actually a particle. The reason they could conclude this was because increasing the intensity of the light in the experiment did not cause any more electrons to be ejected from the metal. However, when ...
Fri Oct 16, 2020 2:35 pm
Forum: Properties of Light
Topic: "in phase" vs "out of phase" [ENDORSED]
Replies: 4
Views: 67

### Re: "in phase" vs "out of phase"[ENDORSED]

Two waves are in phase with each other if their peaks line up together and their troughs line up together. This results in constructive interference where when combined, the amplitude of the new wave is larger. Out of phase is the opposite and means that the peaks and troughs of the two waves are mi...
Fri Oct 16, 2020 2:23 pm
Forum: Properties of Electrons
Topic: Wave Properties of Electrons and the DeBroglie Equation Assessment Question 31
Replies: 2
Views: 13

### Re: Wave Properties of Electrons and the DeBroglie Equation Assessment Question 31

The way I would start is by writing down the de Broglie equation: wavelength = h / mv, where m is mass of the electron and v is velocity. The problem gives you both these values, so after plugging them in, you should get your answer. For the second part, I would say yes, since objects with a high ve...
Fri Oct 16, 2020 2:00 pm
Forum: Properties of Electrons
Topic: Using empirical equation for h-atom
Replies: 4
Views: 30

### Re: Using empirical equation for h-atom

The equation he used in the problem was En = (-hR)/n^2, where n is equal to the electron's energy level. Essentially, the fractions of 1/16 and 1/4 come from dividing by n squared (in this case, dividing by 4 squared and 2 squared. Hope this helps!
Tue Oct 13, 2020 1:02 pm
Forum: Photoelectric Effect
Topic: Textbook Question 1B15
Replies: 4
Views: 46

### Re: Textbook Question 1B15

The equation I used to solve this problem was the de Broglie equation: wavelength = h/p or wavelength = h/(mv). For the value of m, I used 9.1 x 10^-31 kg since this is the mass of an electron, and for the value of velocity, I converted the number given in the equation to m/sec by multiplying by 100...
Mon Oct 12, 2020 5:56 pm
Forum: Photoelectric Effect
Topic: Question 28 On Module of Photoelectric Effect
Replies: 5
Views: 36

### Re: Question 28 On Module of Photoelectric Effect

So to start off you have to use the formula for kinetic energy which is Ek = (1/2)mv^2 where m is mass and v is velocity. The problem gives you the value of velocity, so you then need to know what value to use for m. In this case, m would be the mass of an electron which is 9.1 x 10^-31. (This is ju...
Tue Oct 06, 2020 10:28 pm
Forum: Balancing Chemical Reactions
Topic: Balancing Equations --> #18 in Module
Replies: 4
Views: 78

### Re: Balancing Equations --> #18 in Module

You’re right, whenever a problem talks about combustion, it implies that you’re adding O2 as a reactant. The same thing applies if a problem uses the term burning.
Tue Oct 06, 2020 10:18 pm
Forum: Molarity, Solutions, Dilutions
Topic: G7 The Fundamentals in Textbook
Replies: 2
Views: 40

### Re: G7 The Fundamentals in Textbook

You’re right! I just took 5.45% of 510 grams to find the mass of KNO3 and subtracted that value from 510 to find the mass of water. Don’t worry, I don’t think there’s any secret trick or formula you missed!
Mon Oct 05, 2020 8:35 pm
Forum: Accuracy, Precision, Mole, Other Definitions
Topic: E.1 Nanotechnology
Replies: 3
Views: 70

### Re: E.1 Nanotechnology

The first thing you’ll want to do is find the diameter of one atom by multiplying the radius of 144pm by 2. To find the total length of the fiber, you’ll want to multiply this diameter by how many total atoms there are in fiber. In this case, there is one mol, so the number of atoms in the fiber is ...
Mon Oct 05, 2020 2:31 pm
Forum: Empirical & Molecular Formulas
Topic: Fundamentals L. 39
Replies: 7
Views: 101

### Fundamentals L. 39

In problem L. 39 it says "A 1.50-g sample of metallic tin was placed in a 26.45-g crucible and heated until all the tin had reacted with the oxygen in air to form an oxide. The crucible and product together were found to weigh 28.35 g. (a) What is the empirical formula of the oxide?" Could...