@APScholar18 - Do you mean Kp? It’s the constant for gases ([products]/[reactants]), so large Kp = large numerator = equilibrium lies to the right.
@glasshours no I mean Ksp the solubility product
@APScholar18 High Ksp means that the ionic compound is relatively soluble in water. If anything, you’ll have to compare Ksp values to determine which substance is most soluble in water. I could see them giving a question like this:
The Ksp of NH4X at YºC is 3. Which of the following diagrams shows the species present in significant concentrations in a 1 M solution of NH4X?
The answer would be the diagram with NH4+ and X-. Since K > 1, the products are favored, making them present in more significant concentrations than NH4X. However, there would probably be some NH4X since 3 is not much greater than 1 (can someone back me up on this please?). They’ll most likely give you a very large Ksp value (probably like 100 or something).
Btw, my MacBook corrected Ksp to Kip at first as well lol
Large Ksp usually equals greater solubility, but you also have to take the number of ions into account.
So if you have XY3 -> X3+ + 3Y-, Ksp = [X3+][Y-]^3 = (x)(3x)^3 = 27x^4. Divide Ksp by 27 and take the fourth root to find the solubility. Usually, the greater the exponents, the smaller the solubility.
No idea what @apactstudent said
Can someone explain me 32 from 2013 practice?
@glasshours so how do you apply that to 28 from 2013?
^Haven’t done 2013 yet, saving it for tomorrow (…it’s 11:55 PM where I am, though, so maybe in an hour? I’ll try to explain it if nobody else does)
Honestly, I really hope there’s at least one FRQ dedicated to acid/base equilibria. It’s the only thing I know. 
Can someone help me with #34 on 2013?
@glasshours seriously? thats the main thing i DON’T know
Do you guys think we’re likely to see much of organic chem on it. I’m talking about naming the stuff? In particular like I’ll be find with a simple propane or whatever, but if they throw a aldehyde or Ester at us I will be so lost
@dsi411 Do we have to do dimensional analysis crap though D:? I can do it, but I write so slow, lol.
@thelinah - Yeah, I reread that textbook chapter at least 4 times out of fear of failing my test in class, then did 200+ problems. I’m pretty sure I actually broke down at one point because I couldn’t understand the logic behind those problems. Now it’s the only thing I actually remember.
@Newdle You definitely want to understand dimensional analysis. It’s super useful if you don’t really know what you’re doing because you can follow the units to the correct answer
@KenJaben Does that mean we have to write everything out on the FRQs, though? I know it’s really useful at times, but I don’t wan to write out certain steps if they’re super-obvious…like 1L/1000mL
@Newdle super obvious things like that can most likely be done mentally and therefore really don’t have to be written down unless you really want to. Personally I don’t think they would mark it wrong unless the end units are incorrect. You DO want to show work for formulas you use though in order to show how you arrived at your answer and one more thing; SIGNIFICANT FIGURES!!!
@Newdle on the scoring guidelines, it seems like they want to see any steps that involve the calculation of moles. If we get a question about electrolysis, I would show each step of the dimensional analysis
@KenJaben @apactstudent K, sounds cool. And… Of course I know we have to use sig digs, silly :)… I’m really expecting electrolysis this year, too, haha… Electrolysis is just an unfavorable galvanic sell reaction, though? We might have to I=q/t it, but meh, not too much pain (dimensional analysis doesn’t help there, but I guess we’ll probably have to use it then because there are a few equations involved).
hi guys two questions
is writing chemical reactions still included and how do you do question like how many grams of Zn will form if 0.2F is passed through ZnNO3
@APScholar18
Since you never got a good answer…
For #28, you are told that is AgNO3 and NaCl are present. In a double replacement reaction, this would form AgCl and NaNO3. AgCl is insoluble and NaNO3 is soluble (all nitrates, sodium, potassium, ammonium salts are solube). So, AgCl(s), a precipitate forms from Ag+ and Cl- while Na+ and NO3- ions remain entirely in the solution (Can’t be B).
The question, however, is also testing you on whether you know to what extent AgCl precipitates out.
The Ksp value of a salt is a product of the concentrations of its ions’ when it is in solution. So, low Ksp = low ion concentration = low solubility. AgCl has a low Ksp meaning that the concentration of its ions is insignificant. Because of this, for every 1 Ag+ ion, 1 Cl- ion is consumed to precipitate out AgCl to near completion(“near completion” is what makes the ion concentration of Ag+ and Cl- low)
The question also emphasizes the fact that there is EXCESS NaCl. Ag+ reacts with Cl- 1:1, so Ag+ is the limiting reactant. You shouldn’t expect to see Ag+ ions in solution because of this (Can’t be D). You should, however, expect Cl- ions to be present since it is in excess (Can’t be A).
So the answer is C.
For #32, you are told about the initial and final conditions of a system as it reaches equilibrium. You should make an ICE table.
<code>PCl5`````````PCl3Cl2
Initial1 atm0``````0
Change`-x+x```````````+x
Equilibrium1-xx```````x
You know that the total atm pressure at equilibrium is 1.4atm. Total pressure is equal to the sum of the partial pressures, so
(1-x) + x + x = 1.4
x = .4
Since now you know the concentrations, you can calculate the Kp value to be (.4*.4)/.6 = .16/.6.
Numerator is less than denominator so Kp is less than one.
34 should be treated like a stoich problem.
This question tricks you by asking you abut the reverse reaction. It is saying that the reactant is now KCl and the product is now Cl2. The enthalpy just flips the sign when this happens.
It is now KCL -> K + .5Cl2 H = +437kJ/mol rxn
When .5 mol Cl2 forms, 437 kJ is ABSORBED.
.05 is a tenth of that, so 43.7 should be ABSORBED.