Search found 51 matches
- Sat Mar 13, 2021 5:25 pm
- Forum: Calculating Work of Expansion
- Topic: when is change in internal energy 0
- Replies: 10
- Views: 1938
Re: when is change in internal energy 0
ΔU=0 when q+w=0, or q=-w. In other words, when all the heat gained by the system is expended as work. While this is true in many situations, it isn't guaranteed: For example, in this problem, the work expended should be lower than the heat gained if I did my math right, meaning that q=/=-w, and thus...
- Sat Mar 13, 2021 4:56 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Bomb Calorimeter
- Replies: 14
- Views: 4923
Re: Bomb Calorimeter
I've heard about an open, closed, and isolated system but what exactly is a perfect system? What characterizes such a system and are there any examples besides a bomb calorimeter? I would like to know as well, though I'm pretty sure that it's just another name for an isolated system. I checked thro...
- Sat Mar 13, 2021 4:39 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Acids & Bases
- Replies: 4
- Views: 383
Re: Acids & Bases
Someone else shared a mnemonic for strong bases as well: "For strong bases, it's Lily and Nate RoBbed a Bank and Killed a Cessium of Strong Cats" (Li, Na, Rb, Ba, K, Cs, Sr, Ca)". Thanks to another chemistry community post here https://lavelle.chem.ucla.edu/forum/viewtopic.php?p=30443...
- Sat Mar 13, 2021 4:16 pm
- Forum: Student Social/Study Group
- Topic: Final Jitters
- Replies: 457
- Views: 430046
Re: Final Jitters
Does anyone know how to calm down your anxieties before you take this final? Since having test anxiety can be very distracting while you're trying to study or even while you're taking the test itself. Open to any tips:) Does anyone just zone out during the exam or get stuck in a brain fog and just ...
- Sat Mar 13, 2021 4:04 pm
- Forum: General Rate Laws
- Topic: Units for k
- Replies: 11
- Views: 793
Re: Units for k
If you wanted a equation for it, the units would be in 1/(Mz-1*s), where z is the order for the reaction.
- Sun Mar 07, 2021 9:43 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Sapling Question
- Replies: 3
- Views: 188
Re: Sapling Question
In a similar vein, for the second, since you multiply [A] by 3 and [B] by 1/2, and [B] is squared so it's by 1/4, you end up multiplying it with 3/4.
- Sun Mar 07, 2021 8:33 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Sapling #7 Weeks 9/10
- Replies: 3
- Views: 239
Re: Sapling #7 Weeks 9/10
What you'll want to do to start this out is try and figure out the order of each reactant, so you'd find reactions where the concentrations of everything else are the same except the one you're looking for, and see how the change in concentration affects the rate. For example, if you wanted to find ...
- Sun Mar 07, 2021 7:40 pm
- Forum: General Rate Laws
- Topic: Sapling #8 Week 9/10
- Replies: 7
- Views: 491
Re: Sapling #8 Week 9/10
Since you know it's a zero order reaction, you can use the equation [A]0-[A]i=-kt, where k is the given rate constant and t is the time you're looking for.
- Sun Mar 07, 2021 7:07 pm
- Forum: General Rate Laws
- Topic: sapling #2 week 9/10
- Replies: 7
- Views: 382
Re: sapling #2 week 9/10
Your method should be right, as far as I'm aware. I did the same thing. I guess just double check to make sure everything is correct.
- Sun Mar 07, 2021 6:45 pm
- Forum: General Rate Laws
- Topic: overall order of the rxn (sapling #7)
- Replies: 9
- Views: 679
Re: overall order of the rxn (sapling #7)
How do we know C is the zero order reactant? We know because in the context of the problem, there are two examples where the concentrations of [A] and [B] stay the same and [C] changes, yet their rates do not change. Since the rates do not change, that means that the concentration of [C] cannot aff...
- Sun Feb 28, 2021 8:48 pm
- Forum: Balancing Redox Reactions
- Topic: Balancing Redox Reactions
- Replies: 6
- Views: 404
Re: Balancing Redox Reactions
I think it depends on whether the reaction is basic or acidic. For example, in MnO 2 -> MnO 4 - , the product side has more oxygens, so you'd need to add water on the reactants side to balance it out, and add hydrogen ions on the product side to balance that out. However, that's for an acidic soluti...
- Sun Feb 28, 2021 8:40 pm
- Forum: Balancing Redox Reactions
- Topic: balancing a redox in hw #5
- Replies: 2
- Views: 176
Re: balancing a redox in hw #5
You haven't balanced the electrical charges. The oxidation side has 8 electrons and the reduction side has 6. Multiply the items in the oxidation half reaction by 3 and the items in the reduction half reaction by 4 to balance out the electrical charge.
- Sun Feb 28, 2021 8:11 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Balancing equations
- Replies: 5
- Views: 300
Re: Balancing equations
Whenever one side has more Oxygens/Hydrogens than the other, you'd need to add water and either H+ or OH- to the reaction. Like when going from MnO2 to MnO4-, the extra 2 oxygens have to come from somewhere. So you'd add water to one side to provide the oxygen, and add H+ on the other side to balanc...
- Sun Feb 28, 2021 8:07 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Sapling #16 Wk7/8
- Replies: 8
- Views: 1803
Re: Sapling #16 Wk7/8
For a slightly different perspective, since E=E°-(RT/nF)*ln(Q), we can rewrite it as E=E°-((ln(Q)*RT)/nF). Both n and ln(Q) double, so you end up with E=E°-((2ln(Q)RT)/2nF), and the 2 on each side cancels out to our original equation.
- Sun Feb 28, 2021 7:56 pm
- Forum: Balancing Redox Reactions
- Topic: Balancing Redox Reactions in Basic Conditions
- Replies: 7
- Views: 542
Re: Balancing Redox Reactions in Basic Conditions
To add onto the first responder's process, that's all you have to do for acidic conditions. However, for a reaction in a basic solution, you also have to add OH- ions on both sides to cancel out the H+ ions, which would also cancel out some of the H2O molecules. So in general, the reaction in basic ...
- Sun Feb 21, 2021 10:48 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Oxidation & Reduction
- Replies: 3
- Views: 242
Re: Oxidation & Reduction
I don't know if there's one in the same order as LEO, but if you're fine with a completely different acronym there's OIL RIG, where it's Oxidation Is Losing electrons and Reduction Is Gaining electrons.
- Sun Feb 21, 2021 9:18 pm
- Forum: Balancing Redox Reactions
- Topic: Sapling week 7/8 #2
- Replies: 1
- Views: 169
Re: Sapling week 7/8 #2
The first thing to note is that each of the reactions is a half reaction, so you would want to add them for a full reaction. From there, since you want each side to have an equal charge, you would balance the electrons (and make sure the rest of the equation is still balanced afterward). Hope that h...
- Sun Feb 21, 2021 9:06 pm
- Forum: Balancing Redox Reactions
- Topic: Sapling question 2
- Replies: 3
- Views: 188
Re: Sapling question 2
Are you capitalizing the letters? Sapling can get pretty specific with that, and I noticed that I had to capitalize each element or else nothing would show up.
- Sun Feb 21, 2021 9:04 pm
- Forum: Balancing Redox Reactions
- Topic: Sapling Week 7/8 HW #
- Replies: 1
- Views: 115
Re: Sapling Week 7/8 HW #
For this problem, I'm pretty sure you're able to just balance out the equation like we had done before so that there are equal amounts of all atoms on both sides. You could also check the charges afterward to make sure, but the former method worked for me. If that's not working, can you get more spe...
- Sun Feb 21, 2021 8:39 pm
- Forum: Balancing Redox Reactions
- Topic: Oxidation numbers
- Replies: 4
- Views: 286
Re: Oxidation numbers
I don't think you need to memorize all of them, but there are some common elements to keep in mind. Oxygen is usually -2, Hydrogen is usually +1, Alkali earth metals are usually +2. I think also alkali metals are +1 and halogens are -1. Once you do have them, then it's usually just calculating it ou...
- Sun Feb 14, 2021 10:37 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: 1499
Re: Is it necessary to memorize Cp,m or Cv,m values? [ENDORSED]
Is there a difference when a question says Cp,m/Cv,m or Cp/Cv? I saw the book used both when explaining this concept A Cp,m or Cv,m is just the pressure/volume molar heat capacity, or the heat capacity for a single mol. You'd have to convert one value to the other by multiplying by the number of mo...
- Sun Feb 14, 2021 10:26 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Delta H and Delta S
- Replies: 2
- Views: 676
Re: Delta H and Delta S
Delta H and Delta S are usually related when that if one is increasing, the other is probably increasing as well. Delta H is just the measure of heat in the system, so if it's positive, then heat is going into the system. Delta S is the measure of entropy/the number of positions, and that usually in...
- Sun Feb 14, 2021 10:20 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Why is deltaU = 0?
- Replies: 5
- Views: 305
Re: Why is deltaU = 0?
deltaU would be 0, but q and w wouldn't have to be zero. More generally, it ends up that q=-w, or that any heat gained by the system is expended as work, resulting in a net energy change of 0.
Isothermal just means that the temperature remains constant.
Isothermal just means that the temperature remains constant.
- Sun Feb 14, 2021 9:56 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Determining the Sign of S
- Replies: 8
- Views: 489
Re: Determining the Sign of S
You can also intuitively think about it by knowing that by being positive, it would generally involve gaining heat and go from solid->liquid->gas, as well as being vice versa for negative. However, that should only be to make sure. Whatever you calculate should give the right sign anyways, due to th...
- Sun Feb 14, 2021 9:18 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Sapling 3
- Replies: 13
- Views: 633
Re: Sapling 3
I don't know why I was doubting myself thinking there had to be something in the middle columns too. Thank you! could you explain why nothing goes in the middle columns? is it impossible for both dH to be negative and dS to be positive (or vice versa)? thank you! Nothing goes in the middle columns ...
- Sun Feb 07, 2021 11:03 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Sapling Week 3/4 #20
- Replies: 3
- Views: 148
Re: Sapling Week 3/4 #20
In question 20 specifically, q=deltaU because it is at a constant volume, and so no work is being done. If w=0 and deltaU=q+w then deltaU=q.
- Sun Feb 07, 2021 10:56 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Sapling #12 Week 3/4
- Replies: 5
- Views: 259
Re: Sapling #12 Week 3/4
Hi,
You would find the calorimeter constant by the equation C=q/deltaT. And to find q, you would just convert the mass of the compound to mols and then multiply with the heat of combustion. Hope that helped!
You would find the calorimeter constant by the equation C=q/deltaT. And to find q, you would just convert the mass of the compound to mols and then multiply with the heat of combustion. Hope that helped!
- Sun Feb 07, 2021 10:45 pm
- Forum: Phase Changes & Related Calculations
- Topic: Knowing reversible and irreversible
- Replies: 9
- Views: 451
Re: Knowing reversible and irreversible
I know that reversible equations are also known as isothermic, and that irreversible ones are used when the system is being heated over a constant pressure.
- Sun Feb 07, 2021 9:46 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Sapling #14 Week 3/4
- Replies: 7
- Views: 398
Re: Sapling #14 Week 3/4
You might need to multiply the -pressure times volume by 101 J to convert to joules. I was having trouble with this as well and this solved it, but why does this work? I don't really get where the conversion comes in. EDIT: Well seems the poster above me preemptively sniped my question with an answ...
- Sun Feb 07, 2021 9:10 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Sapling Week 3/4 #11
- Replies: 5
- Views: 281
Re: Sapling Week 3/4 #11
I actually figured out this problem by remembering that grams can be roughly calculated as 1 g/ml, however I am confused as to why I don't need to include delta H fus in this equation, unlike previous sapling problems. Could someone explain to me when delta H fus is necessary? Thanks! Delta Hfus is...
- Sun Jan 31, 2021 11:16 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Reverse direction (sapling q.5)
- Replies: 7
- Views: 373
Re: Reverse direction (sapling q.5)
For this specific example, we were given MCl3(s) ---> MCl3(aq) ΔH4 = -388.0, however when you break down the reaction we were looking at MCl3 changed from (aq) to (s), therefore we need to use the reverse of the reaction enthalpy we were given for that change. Hence you use -2 instead of just 2. Do...
- Sun Jan 31, 2021 10:55 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Closed vs Isolated Systems
- Replies: 4
- Views: 243
Re: Closed vs Isolated Systems
A closed system is just a system where matter can't escape (like a sealed container), but an isolated system is a system where energy cannot escape (like a bomb calorimeter... Theroretically, anyways). In the closed system, let's say you're holding a sealed container of water, none of the water can ...
- Sun Jan 31, 2021 10:47 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Pressure
- Replies: 7
- Views: 397
Re: Pressure
Constant pressure doesn't mean no pressure! It just means that there is no change in final and initial pressure. It could be both 0 or a nonnegative number. Why a nonnegative number specifically? Technically, pressure can never be less than 0, and technically it can. It depends on what your referen...
- Sun Jan 31, 2021 10:45 pm
- Forum: Phase Changes & Related Calculations
- Topic: X2 vs 2X
- Replies: 14
- Views: 844
Re: X2 vs 2X
I believe that you would figure out this problem by utilizing the fact that enthalpy is the (change in energy of bonds broken) - (change in energy for bonds formed). In this reaction, the x2 is being broken and there are no bonds being formed: So it would be (bond energy of breaking X2) - 0 which wo...
- Sun Jan 31, 2021 10:33 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Pressure
- Replies: 7
- Views: 397
Re: Pressure
Jamie2002 wrote:Constant pressure doesn't mean no pressure! It just means that there is no change in final and initial pressure. It could be both 0 or a nonnegative number.
Why a nonnegative number specifically?
- Sun Jan 31, 2021 10:16 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: When to convert mass of a substance given to moles
- Replies: 5
- Views: 270
Re: When to convert mass of a substance given to moles
To add on, you would specifically convert whenever your needed value is measured in moles. Since delta h is commonly measured in kJ/mol, it's likely that you would need to convert the mass to moles. However, in the future, if theoretical problems involve us converting to some other value measured in...
- Sun Jan 24, 2021 11:05 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Sapling Number 5
- Replies: 18
- Views: 737
Re: Sapling Number 5
Kandyce Lance 3E wrote:For this problem, are [OH-] and [BH+] interchangeable ?
Yes, since it's a monoprotic reaction, both values taken in this problem are the same. Although it would probably be good form to use [OH-] and [BH+] properly, just to lessen the chances of messing up in the future when they're not.
- Sun Jan 24, 2021 11:00 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: When to use ICE box
- Replies: 23
- Views: 1037
Re: When to use ICE box
It's also worth noting that ICE boxes can be used to solve equilibrium problems involving pressure as well, not just concentration. One of the problems in the previous Sapling HWs had that.
- Sun Jan 24, 2021 10:57 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Pka, Ph, charged and neutral species
- Replies: 7
- Views: 437
Re: Pka, Ph, charged and neutral species
You know, that makes me curious. If it was pOH and pKb instead, would it still be the same thing, just reversed? As in bases have charged species dominant for pOH>pKb and neutral species dominant for pOH<pKb, and vice versa for acids. Intuition tells me yes, but I'm still not sure.
- Sun Jan 24, 2021 10:30 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: sapling week 2 #5
- Replies: 15
- Views: 579
Re: sapling week 2 #5
I had a lot of issues with this problem too. I think the biggest thing I needed to remember is that you don't just need the concentrations for [B], you need the concentrations for [B]initial. [B]initial is just [B]+[OH-], and then you'd use that value to find the percent ionization.
- Sun Jan 24, 2021 10:25 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Sapling Homework Week 2
- Replies: 5
- Views: 363
Re: Sapling Homework Week 2
If Ka/Kb significantly favors one over the over (As in the value is k<10-4 or k>10^4), and the percent ionization is less than 5%, then it is safe to assume that x is neglectable. Mathematically, it means that it's small enough that you would essentially have the same concentration, and those ratios...
- Mon Jan 18, 2021 12:04 am
- Forum: Phase Changes & Related Calculations
- Topic: Usage of Ph with acids and bases
- Replies: 6
- Views: 282
Re: Usage of Ph with acids and bases
pH is just a convenient way to notate it. Saying that the pH is 2.15 is much more convenient than saying the concentration is 0.0070129. Because it's a negative log as well, we can quantitatively tell how strong an acid/base is. The inverse of -log(concentration) is 10^(-pH). If it is a strong acid ...
- Sun Jan 17, 2021 11:54 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Ka and Kb relationship
- Replies: 5
- Views: 239
Re: Ka and Kb relationship
Mathematically, Kw=Ka*Kb=10^-14. So as one goes down, the other increases.
Conceptually, I'd imagine it has to do with how acids and bases are related. A strong acid dissociates into a weak base, for example. It's just due to how they're related. I hoped that helped.
Conceptually, I'd imagine it has to do with how acids and bases are related. A strong acid dissociates into a weak base, for example. It's just due to how they're related. I hoped that helped.
- Sun Jan 17, 2021 11:46 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Sapling #5
- Replies: 8
- Views: 399
Re: Sapling #5
You would use some rules of combining chemical equations to combine a few of the 4 listed into the top equation. The specific rules are -Switching the products and reactants [R] <-> [P] into [P] <-> [R] would make the new Kc value equal 1/Kc -Multiplying a chemical equation n([R] <-> [P]) = n[R] <->...
- Sun Jan 17, 2021 11:30 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: +/- x in the ice box
- Replies: 4
- Views: 329
Re: +/- x in the ice box
You know depending on which side the chemical equation is shifting. If you add more reactants initially, then the change for molecules on the left side would be -x (with stochiometric coefficients). The change on the products side would be +x (again with stochiometric coefficents). Same for if they ...
- Sun Jan 17, 2021 11:19 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: ICE Box
- Replies: 10
- Views: 564
Re: ICE Box
You know depending on the reaction equation. If the initial values are on the left side of the equation/you add reactants initially, then the change would be -x for all the molecules on that side (with stoichiometric coefficients), with the right side/products being +x (again with stochiometric coef...
- Sun Jan 10, 2021 11:50 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Heterogenous Equilibriums
- Replies: 4
- Views: 217
Re: Heterogenous Equilibriums
SamayaJoshi1A wrote:Can someone explain what the term (aq) means? I know that it is aqueous, but how does that differ from a liquid? Is it because its a solution?
I believe the term aqueous is specifically referring to something dissolved in another liquid.
- Sun Jan 10, 2021 11:47 pm
- Forum: Ideal Gases
- Topic: Inert Gases
- Replies: 11
- Views: 364
Re: Inert Gases
I’m confused on how this situation presented in 5J of the book can be true if adding gases doesn’t change pressure which then doesn’t effect concentration: “Suppose that the ammonia synthesis reaction, reaction A, has reached equilibrium. Now suppose that more hydrogen gas is pumped in. According t...
- Sun Jan 10, 2021 11:39 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Rice method correlation [ENDORSED]
- Replies: 4
- Views: 246
Re: Rice method correlation [ENDORSED]
I haven't used it, but from what I could tell from searching it up, it is essentially the same thing. Just differences in naming.
- Sun Jan 10, 2021 11:25 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: ICE Tables
- Replies: 11
- Views: 764
Re: ICE Tables
In a general sense, whenever you have an equation that is going to equilibrium, it helps to make an ICE table. It was mostly used for concentrations, but it was also used for pressure in the homework, so it's not strictly limited to finding concentrations.
Edit: Got more specific.
Edit: Got more specific.
- Sun Jan 10, 2021 11:13 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Sapling #6
- Replies: 2
- Views: 134
Re: Sapling #6
I'm pretty sure the temperature is only there because any reaction only changes it's Kc value at a different temperature. Since the temperature is listed as 500 degrees Celsius both times, it's another way of saying you can use the listed Kc value for the problem. It doesn't factor into the calculat...