okay haha thanks guys! crossing my fingers that buffers stay in the mc
ohh there’s a 2014 test out??!
Lol chris work on your math skills, mc and crq are worth the same amount so 50% on each is a 50% in total brah
78 composite is 78/150 which is 52% the numbers posted above were in percentages, not composite scores
I have a question on net ionic equations for titrations.
If you have NaOH and HC2H3O2
Is the net ionic
NaOH + HC2H3O2 —> NaC2H3O2 + H2O
Then
C2H3O2- + H2O --> OH- + HC2H3O3
Yea Na is not in the net ionic, the net ionic is oh- + ch3cooh --> ch3coo- + h2o
@Chris360911 what are you talking about how in the world is 50% a 5?
@glasshours
H2 is consumed with a ratio of 2:1 compared to CO2. For any reduction in CO, the reduction is twice for H2.
It was easier for me to do this problem if I used dummy values (based on the Kc value, they are way off, but it demonstrates the concentrations relative to each other while the system establishes equilibrium).
If .2 mol H2 were used, [H2] would be .8, [CO] would be .9, and [CH3OH] would be .1
checking this with the choices, the only one that aligns with this is B.
^ I had trouble with that question too , how did you do the second part of the question? About the total pressure at constant temperature, isn’t it just P1V1=P2V2?
Are you guys talking about mc or frq
40/60 on the MC, 27 on last year’s FRQ… what does that translate to?
No I meant the equation after “Then” :).
So the net has the oh- as the reactant but for calculations purposes you flip it?
OK What are the chances buffers show up on the frq bc I DO NOT want them too. Also, I hate those titration curve questions where it asks you which concentration is the highest at this point on the graph.
also, how much of the MC is calculations… i’m thinking the majority is theoretical
Zep thats a 4, I think it’s mostly concept based youre right
the hardest concept for me is electrochemistry… any tips? last year’s FRQ had some weird questions with common-sense answers but i got them wrong
Yea that’s a low 4
Just remember AN OX and RED CAT (anode is oxidation, reduction is cathode) and that for galvanic, the lower reduction potential oxidizes. Also, for electrolytic cells, for me all I do is find the galvanic version but then flip it to find the electrolytic version.
@sunsh0wers
I had trouble with that part too. I’ll give what I think is happening (im not 100% sure)…
What I think is that you use P1V1 = P2V2 to find the INITIAL total pressure. The system is now at that new pressure without having established equilibrium yet.
Since pressure increases, equilibrium shifts to the right (least number of moles)
When the system moves towards equilibrium, 1 mol of CO and 2 mol of H2 are consumed for 1 mol of CH3OH, reducing the total number of moles (change in moles is -2). Based on this, it can not be C or D.
But, since the K value of the reaction was so high, the change that results from decreasing the partial pressures of reactants is so minimal that it does not bring it below 1.2 atm again. So, the answer is B.
yeah, there’s my half-assed attempt at explaining it. again, im not completely sure.
Just a recap, the metal that loses the electrons is being oxidized and is the anode, gains electrons is cathode which is being reduced. The cathode gains electrons which turns the ions in its solution into the metal itself, causing the metal’s mass to increase. On the other side, the anode loses its electrons causing the metal to turn into an ion which goes into the solution.
Also voltage doesnt depend on moles
Thats about it
Taw how did you find the test?