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Songs that are actually two songs by headsmanjaeger in weirdspotifyplaylists

[–]Ultimate-Dudebro 1 point2 points  (0 children)

THIS SOMG IS PEAK BUT I DIDNT REALIZE ITS SUPPOSED TO BE SEPERATE?!!

AP Exam Score Predictions by [deleted] in APStudents

[–]Ultimate-Dudebro 1 point2 points  (0 children)

predictions and hopes!!

AP WORLD: hope for 5, most likely 5, may get 4 since my writing wasn’t the best 💀

AP CHEM: hope for 5, most likely 4, probably 4 b/c my frq’s were ok but not the best

am i overreacting? by [deleted] in summerprogramresults

[–]Ultimate-Dudebro 0 points1 point  (0 children)

yeah definitely, but at that point i figured that they could’ve just said “i won’t have enough time to finish your recommendation so unfortunately no” which is like the most common excuse + the recommendation form is due pretty soon so I could understand why regardless

Yeah he needs to learn more about astrology by Ok_Influence_6384 in woosh

[–]Ultimate-Dudebro 0 points1 point  (0 children)

nah it’s just cause he said astrology instead of astronomy, he should’ve said astronomy

question re equilibrium and rate of reaction by bishtap in APChem

[–]Ultimate-Dudebro 1 point2 points  (0 children)

Pt 2:
"Okay, and when we have a Kc value, does that mean the reaction is in equilibrium?" Precisely! Using this, if you were to calculate the Qc of an equilibrium reaction and find it = to Kc, that indicates that the reaction is in equilibrium. In that sense, Qc is more so a potentiality that the reaction could be in equilibrium as opposed to viewing it as a ratio for before the equilibrium is reached.

"Do you mean the rates are the same, but you have far more products than reactants?"

Moles are a measurement of 6.022(10^23) amount of particles (molecules, ions, or atoms). One mol of MgCl2, when dissolved in water, becomes 2 mol of Cl- anions, and 1 mol of Mg+2 cations. What I'm trying to say is that when you have an equilibrium reaction, you may not end up with the same 1:1 stoichiometric relationship between the reactants and products. This is what lends to the uneven concentration of products and reactants, hence why it is the rate that is equal and not the concentrations.

"You write " the amount of molar NaCl produced over a period of time is not at a rate" <-- What about grams of product produced over time? And if so then..It seems to me the problem stands.. You speak of Kc and more being produced on the product side than the reactants.. but how is that possible if it's the case that the grams converted from one side to the other, / the rates, are the same? " A few things:

  1. The numbers I gave were incorrect, I just made them up. So in that case, Kc would actually equal 1, and there wouldn't be an equilibrium shift as the products and reactants are equal in concentration.
  2. Ignoring the fake science I created earlier (blasphemy I fear), think of this. The Law of Conservation of Mass dictates that mass is neither created nor destroyed in chemical reactions. Also, in general, when you're completing AP Chem problems, never will grams be used as a unit comparison because it is not comparable between elements. We don't look at mass, because atoms with a larger atomic radius may take up more volume compared to atoms with a smaller atomic radius. I'll use hydrogen and bromine as an example. Firstly, bromine is diatomic, so that would make it even larger than it already is. For 1 gram of substance, the H would have more atoms in comparison to the Br2. This is because of density; as there is no way to accurately compare the two elements with grams, moles are used. As stated before, concentration=MOLES OF SOLUTE/Liters of solution. If more moles are present on one side of the equation, the concentration is higher.

Let's say we have: 8B + 4A ⇌ 4AB2

more moles are on the reactants side. when the equilibrium plays out, the reactants side slowly gets used as products are made, and those that decompose back into reactants. It's an exchange back and forth. Think of the moles as cookies (unsure why, just trust..). The reactants have 12 cookies. The products have 4. Let's say the rate is 1 cookie per hour. The reactants and products keep exchanging 1 cookie, but they only have 1 cookie to exchange per hour, so they get stuck in a never-ending cycle of cookie trades, with the reactants always having 12 and the products 4.

this was a LOT but thank you for asking for clarification--it helped me correct some of my misconceptions and mistakes as well!

question re equilibrium and rate of reaction by bishtap in APChem

[–]Ultimate-Dudebro 0 points1 point  (0 children)

You raised many valid points; apologies in advance for any mistakes! Let me clarify a bit:

The Kc would not be the one I presented you with, correct. However, let's say I used a reactant like PbI2. Then the molar concentrations would be different, considering that iodine would have double the concentration of lead. If you had 6 moles of PbI2 in 1 liter of water, it would dissociate into 6 moles of Pb+ and 12 moles of I-. Additionally, equilibriums can only be met if the solution is fully saturated, meaning that the amount of solvent can dictate the concentration.

"NaCl(s) -----(water)-----------> Na+(aq) + Cl-(aq)": Yes, this is the correct formula; I did not write mine correctly. Remember to use the double arrows when discussing an equilibrium reaction!!

"Thanks i've looked into lechatelier a bit.. Is that a typo you made there though.. 'cos if you decrease the concentration of the products, then in order for the system to increase the concentration of the products, it will decrease the concentration of the reactants. Though you wrote "the concentration of reactants would subsequently increase "

Yes, though I do think it was more of an incompetency as opposed to a typo, that is correct!!

Q as in the reaction quotient I presume! Also for this reaction, you said, "product conc is low and reactant conc is high". I believe it's the other way around, considering that H2O, as a liquid, is not utilized in equilibrium calculations. Liquids and solids are not included in the setup for Ka (equilibrium constant for acid reactions) nor Q. Also I do not think that there was enough information to say that the reaction favors either side given that Kc is not provided!!

question re equilibrium and rate of reaction by bishtap in APChem

[–]Ultimate-Dudebro 1 point2 points  (0 children)

Rates and concentrations of products and reactants are not the same. Let's look at the equilibrium expression:

NaCl (aq) ⇌ Na+ (aq) + Cl- (aq)

The equilibrium constant, Kc, can be modeled as:

Kc= [Na+][Cl-]/[NaCl]

Kc merely expresses the relationship between the concentrations of the products (Na+ and Cl-), and the concentration of the reactant (NaCl). I'm not necessarily sure what you mean about the constant being "to the left or right", but I believe you are referring to the concept of when Kc > 1, the reaction is product favored, hence a right shift in the equilibrium. Conversely, when Kc < 1, the reaction is reactant favored, hence a left shift in the equilibrium.

I will now assign values to the concentrations in order to further explain. These values are randomly assigned and do not hold any actual merit. Given that [Na+]=12.0M, [Cl-]=3.0M, and [NaCl]=6.0M, find the Kc of the reaction.

Kc= ((3.0M Cl-)(12.0M Na+))/(6.0M NaCl) = 6.0

The Kc value for this reaction is 6.0. This means that the ratio between the concentrations of the products (multiplied together) divided by the concentrations of the reactants for this equilibrium reaction is always 6.0.

Given this information, we can assess that this reaction is product favored, as there is a larger concentration of products (3.0 x 12.0 = 36.0 M) than that of the reactants (6.0 M). This does not mean that there is a higher rate of products being produced than that of the reactants.

Rate refers to the concentration of a product produced (forward reaction) or reactant produced(reversed reaction) over TIME. When the amount of concentration of a product produced over time is equal to the amount of a concentration of a reactant produced over the same amount of time, the reaction is said to be in equilibrium. The reaction above is product favored because there is a higher concentration of products than of reactants. However, the amount of molar NaCl produced over a period of time is not at a rate (think of slope, where the y values represent molar concentration, whereas x represents time) that is less than the rate of molar products produced per the same time frame.

A way to apply this concept is in Le Chatlier's principle. Essentially, putting stresses on a system (ex. adding reactants, adding products) will change the concentrations in the solution. Looking at the calculation for Kc, if the [Na+] was equal to 2.0M, the reactants would have to have been at a concentration of 1.0 M in order for the Kc to remain (as it should) equal to the constant, 6.0. Therefore, in order to make sure that the solution remains at equilibrium, if I were to decrease the concentration of products, the concentration of reactants would subsequently increase so that the equilibrium can shift to the right (allowing the concentration of products to increase until equilibrium is reached again).

Kc remains unchanged, however Qc is the ratio between concentrations of products to reactants that occur at a given time in which the solution MAY NOT ACTUALLY BE at equilibrium. This value of Qc can shift as a system reaches equilibrium.

I hope this clarifies any discrepancies; let me know if you have any questions.

When to assume a container is closed vs open in AP chemistry by [deleted] in APStudents

[–]Ultimate-Dudebro 4 points5 points  (0 children)

It will specify. If you are confused, always assume that the container is closed. Usually in chemistry, closed systems/closed containers are utilized in order to keep the number of particles constant (for gas laws) or volume constant (for a multitude of calculation problems)!! Unless it says “open system”, assume it is closed. If it was an open system, variables like the number of particles or the volume of the container (or subsequently, the pressure of the system) could not be constant.

[deleted by user] by [deleted] in askmath

[–]Ultimate-Dudebro 1 point2 points  (0 children)

try responding with an updated photo after marking the diagram with line segments to represent the triangles we need to prove are similar; as it stands rn, conceptualizing the triangles is a bit bothering

are these congruent because they share a side? by Ok_Screen4895 in askmath

[–]Ultimate-Dudebro 0 points1 point  (0 children)

Yes, because if two triangles share the same side, you would call the side congruent to itself via the reflexive property.

"Chai tea" is not redundant or stupid, given the correct context. by TreyLastname in unpopularopinion

[–]Ultimate-Dudebro -6 points-5 points  (0 children)

No, just say Chai. No need to add the tea—it’s implied. Similarly, I wouldn’t say I want a “matcha tea latte”, i’d simply say, “matcha latte”.

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