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Thank you for your video, it was really helpful. The Physical Chemistry book by Peter Atkins (11 ed) brings this problem. However, it considers the transition to be from +3 to +4 states, so the answer is wrong. Anyway, thanks again!
@LuxersMultiverse2090 I hear you, and it seems there are way too many examples of it. However, sometimes I think that the way it's used --by the many different actors-- is a much bigger problem than technology itself. Cheers to a healthy, exciting, and productive use of technology~! 👍🏼🥂
@@chocolate_3867 Sometimes , I also look at the motivation of technological inventions. How can we be sure that the technology we invent goes against the natural order or simply hinders a truth from us.
@LuxersMultiverse2090 Agree, here edited to reflect that: I hear you, and it seems there are way too many examples of it. However, sometimes I think that the way it's *conceived, created, implemented, and used* (by the many different actors) is a much bigger problem than technology itself. Cheers to a *responsible, ethical, healthy, exciting, and productive* technology in our lives~! 👍🏼🥂 (Wow, the wishlist grew quite a bit)
@RexAlfieLee Well, I can't speak for her, but I guess that, to her point (in the quote): She fearlessly worked to understand the unknown... until the end. Thanks for the comment. Cheers~!
I can see how the notation might be confusing and potentially frustrating. Although it's pretty much standard, and this notation aims to emphasise the fact that each function is dependent on **only one** of the corresponding variables. Additionally, these are just labels; when one understands that, everything becomes easier to follow. For instance, when one talks about the natural variables of thermodynamic potentials, the nomenclature for "function of" is **not** "f" but it rather seems redundant for newcomers to the field. Like this, The Gibbs energy: G = G(T, p, N) Rather than, say: G = f(T, p, N). Anyway, thanks for your comment. But I have to say, I would have appreciated constructive criticism to make things better, instead. In this case, for example, adding the why for you this merits the characterisation you made would have been quite useful, for example, in pointing out the source of the issue to substantially address it. Cheers 👍🏼
Very well explained, and your answer to @mateihurba8580 clarify more. One clarification i need if possible: σ=5.67x10^-8 W/(m^2.K^4) without steradian comes from that formula with 2 at numerator, and is from a Hemisphere blackbody (what i undestand from 2 sites: spectralcalc radiance: integrating planck equation and NASA TE Michels planck functions and integrals). Is σ or your α constants valids only to Hemispheres? Like to a sphere we need to multiply by 2? Because the Sun sphere is widely stated with a Power of 3.8x10^26 W , using P=4πr^2σT^4 gives that also (T=5780 , r=700Mm), but maybe is wrong by a factor of 2, and is really 7.8x10^26W, by that fórmulas. But truly is 3.8x10^26W, by the independent measure of 1360W/m^2, at the Earth distance of 150MKm. I cannot understand why they talk to hemisphere.
Hi, Thanks for your comment, the quite interesting question, and the reading suggestions~! To your question. It is related to what my other answer discussed before: what quantity are you considering, spectral {intensity, flux, energy density}. One way to see it is that we can express results irrespectively of the size or amount of a sample. Say for example, density of water: It doesn't matter if it is just a glass or a whole pool. The idea is that you characterise it as how much mass is contained per unit volume. Similarly, here we get the spectral energy density (radiant energy contained in the volume of the cavity per unit wavelength, u_{s}=u/dλ). It does not have the square-radians because, for a volume element (dV), it's pointless to relate the energy density to a solid angle (Ω = A/r²), instead, you use the whole volume (a full sphere, dΩ = 4π). This is where the factor of 4π comes from in the relation posted in my previous comment: u_{s}= u/dλ = (B_{s}/c) dΩ = (4π/c) · B_{s} One gets the *total* energy density inside the cavity of a black body by *integrating* the expressions for u_{s} over the *entire* wavelength or frequency range (as in the video). What do we get? That this total energy density depends only on T~! In any event, the hemisphere comes from the concept of a blackbody itself: an opening (leading to a cavity) that absorbs incident radiation and thus emits radiation. Note that this radiation propagates into "half-space." One must consider a surface element (dA) corresponding to the area of the opening, and to measure the total emitted radiant power coming off from it one must sample over the entire "half-space" (2𝛑, hemisphere). This is because our measurement is affected by the orientation of the detector and its "coverage" (this is where the solid angle, Ω = A/r², and its dΩ are relevant). Finally, the reason the Spectral intensity I_{s} and spectral flux B_{s} differ only by the factor π (not by 2π) comes from the integrations carried out. Actually, check out again the first reference you mentioned (integrating the Planck eqn). There, it mentions the "cosθ reduction from Lambert’s cosine law." and you can see how it works. Hope this helps~!
Hi, Thanks for your comment and question. Depending on your version of Excel, please make sure to select the whole array of cells where the results will show. Then, enter the formula, and finally use the key combinations, Windows: CTRL+SHIFT+ENTER Or Mac: COMMAND + SHIFT + RETURN. I think more recent Excel versions will print all the data in all the corresponding cells "automatically', but again, it depends on the version. Also, do not forget to enter the options False, False accordingly, after the x and y values. Let me know if that works~!
@thanhloannguyenha102 Well, if you already are using LINEST, then you should already have the values you need in that array. You can simply refer to the cell where the value you need is printed. You can see in the video what each of the numbers in the array of results for LINEST mean, check it out. Also, you could directly use the functions below to get parameters of the linear fitting: INTERCEPT(known_y's, known_x's) SLOPE(known_y's, known_x's) Cheers~!
Hi, Thanks for your question / comment. I suggest you look into: Shoemaker, D. P., Garland, C. W., and Nibler, J. W. Experiments in Physical Chemistry, 6th ed., McGraw-Hill, NY, 1996, Chapter XIV, Experiment 34 For information on a modified version of the model. That may help with your question. Best~
Thanks a lot for your comment, and your question~! I'm glad your question was answered in the comments already, but thanks for mentioning it. That way, I can realise how important certain details are --as future reference. Happy Learning~!
Hi, Thanks a lot for your question~! It actually goes back to the last few seconds on the video where I ponder about the relationship of the "\alpha" α (radiation constant) I wrote and the Stefan-Boltzman constant, "\sigma" σ. You're not wrong, but we must clarify that there are various forms to write Planck's Radiation Law, according to the quantities considered. Commonly, we can express the radiation emitted by blackbodies using, ● (A) intensity per wavelength (or frequency): distribution of spectral intensity I_{s}. ● (B) intensity per solid angle: specific I_{s} or a.k.a. spectral flux B_{s}. ● (C) spectral energy density: E_{s} -also written as u_{s}- used in this video. Additionally, note that each of these can be put in terms of the radiation's wavelength (\lambda, λ) or frequency ( u, ν). So, yeah, there are several equivalent expressions out there. That being said, and without going into details (too long for a comment here, but I invite you to research about it), I leave you with the relationships among these quantities. That ultimately answers your question, and the question that I pose at the end of the video too: ☆ I_{s} = π · B_{s} ☆ u_{s} = (4π/c) · B_{s} ☆ u_{s} = (4/c) · I_{s} Where c is the speed of light. You can see from those expressions that likely you're referring to B_{s}(λ), whereas I presented a discussion with u_{s}(λ). ○ So, how would you write α in terms of σ? Happy Learning~!
Ah, that makes complete sense! I was indeed referring to the intensity per solid angle... oops. I'll definitely do some research on the different, albeit similar, mentioned quantities. I'm sure they come in handy in other problems. Thank you for the thorough answer! @@chocolate_3867
Hi, That should come about, P(total) ~ 11.16 %. Compare with the ground state probability: Any guess on what would be for the second excited state? Now, think about what the trend for total probability would look like as the system goes to higher and higher energy states, why? Happy Learning~!
HI, thanks for your comment. May I ask what is causing the problem? * Is it the pace or speed? You could slow down the video using settings. * Is it something that can be helped with turning on Closed Caption in settings? * Is it something related to the details of the calculations that are not clear enough? ... Something else? Cheers~
fascinating. What if Planck was covering up the noncommutative calculus truth of reality? See Professor Basil J. Hiley for details. Wonder how that would change this analysis.
@ma.crizellefrancisco2223 Thanks for the suggestion. I just had some time to quickly make a video for the solution to that problem: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-5S3CFoVW8NQ.html Happy Learning~! :)
5 mM (milli-molar) is your concentration. 5 mM = 0.005 M = 5×10^-3 M Not sure what the rest of your data looks like though. But if that is you initial concentration, you have something to work with. Best,
@@chocolate_3867 I have to analysis cyclic voltammetry data and for that I have to plot this graph but you take different Concentration but I have exact concentration 5 millimolar ( that is the limit of detection value ) so how I can plot this graph ?
@@Mathis-World You probably have graphS of current vs potential. Those give you peaks that would be associated with particular chemical species. Then, following changes in peak currents in Anode over those in the Cathode (or peak potentials), as scan rates vary, would allow you to determine kinetics parameters for your experiment. Note that the peak current heights, and their ratio (A/C), are associated with the depletion of a particular chemical species in your system, i.e. its remaining concentration, from where you can pull [conc] vs time related plots. In case you need it, here are 2 resources that may help: • pubs.rsc.org/en/content/articlehtml/2019/sc/c9sc01545k • pubs.acs.org/doi/10.1021/acs.jchemed.7b00361
Hi, You can think of R^2 as the "chunk" of your data that fits a particular model, e.g., R^2 = 0.98 means that about 98 % of your data can be explained with the model. Now, what value of R^2 is "good", depends on the field of study (social vs physical sciences) and that's something to consider. Also, there are some statistical tools that will allow you to calculate *errors in the parameters of your fitting model (linear equation here). You can check out the function LINEST in Excel; here is one of my videos that shows how to use it just in case is needed ---the video deals with making a Calibration Curve (another Chemistry application of linear regression + std deviation from measurements): ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-MKUngRa8gIg.html Cheers~!
Hi, The equation you're proposing is equivalent to the one showed here: [math] ln (C) = -kt + ln(Co) [/math] Rearrange to: [math] ln (C) - ln(Co) = -kt [/math] Use properties of logarithms: [math] ln (C/Co) = -kt [/math] Also, [math] ln (Co/C) = kt [/math] Now, compare these equations: how will they be different graphically? Hint: look at the slope(s) and intercept(s). Happy learning, cheers~!
Hi, I don't remember exactly, but *Any General Chemistry textbook would have problems of this kind. At the time I made this video, I was using a specific textbook; now I think I shouldn't have included the "Problem Number" thing. Cheers~
Not really sure I understand your question, but in general: Kinetics deals with data you get in lab as you measure how the concentration of a given species changes with time, during a chemical reaction. Now for this video, I took some "toy model" data from a general chemistry book. Any book will have many of these problems.
Thank you so much again for this! Ive been struggling to find tutorials for my Chemical Reaction Engineering subject, so glad i found your channel. Keep up the good work
Short answer: No. If you are seeing those values, then that might be an artefact of the transformation (linearisation) of the original expressions for your pseudo-order kinetics, under certain conditions. My suggestion is to *not use the linearised form, but rather use the original equations to obtain the relevant parameters of your experiments ---this implies running a Non-Linear Fitting... That is indeed a good idea for a possible next post (Thanks for that~!) Best,
It's been a busy season all the way from Summer, but thanks a lot for your suggestion on additional material. It sounds interesting, and I will probably work on it to post.
Good question~! 👍 You could use the General rate law equation: [math] r = k C^n [/math] Using properties of logarithms: [math] log(r) = log(k) + n log(C) [/math] There is your linear equation~! What values would you use? See my video on initial rates method: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-8ZeHIiaCpRk.html ** However, note that in some cases this can be problematic (due to linearisation artefacts) and not the best way to determine a rate order~! Perhaps you should be looking at the use of Non-linear fitting methods. Hope this helps~
Hi Gabrielle, Unfortunately, I don't remember at this moment. But data of this kind is very typical of problems you would find in Any Chemistry textbook with a chapter on Chemical Kinetics. The method described here is general, in that sense. Hope this helps.
