<p>haha so my stupid way of remembering it held up:</p>
<p>the incredible density of the solid pushes the line towards the liquid :P</p>
<p>Nahhh, I would think the same thing. But since KOH is concentrated, it doesn’t ionize in the beginning, and Zn(OH)2 is a precipitate, so it isn’t ionized.</p>
<p>OH concentrated MY B… i suck at 4 lol i average maybe 6-7 pts? but thankfully the rest of the AP usually goes really well for me so i’ve been a safe 5 on all of the practice tests</p>
<p>answers:
a) Na + H2O –> Na+ + OH- + H2</p>
<p>b) H+ + HCO3- –> H2O + CO2 (Part credit for H2CO3)</p>
<p>c) C2H5OH + HCOOH –> HCOOC2H5 + H2O</p>
<p>d) OH-+ Zn(OH)2 –> Zn(OH)42- (or Zn(OH)3- or ZnO22- + H2O)</p>
<p>e) BF3 + NH3 –> BF3NH3</p>
<p>Oh, haha. It’s fine. I didn’t know about that either until my teacher drilled alot of the silly rules into our heads, haha.</p>
<p>Whats the significance of the value E(cell)? How is it effected? What happens when its + or - or 0? </p>
<p>In the equation Zn + Ni2+ -> Zn2+ + Ni E = 0.51 V</p>
<p>If i decrease concentration of Ni what happens to E and why?</p>
<p>How do the practice tests in Princeton Review compare to the real test?</p>
<p>Because I got a raw score of 56 on the MC, and I don’t feel like trying the FRQ just yet.</p>
<p>@miheonigirisan: If the voltage is positive, the reaction is spontaneous, if negative, not spontaneous. Not sure about 0, haha. Doubt they’d make it like that</p>
<p>Not sure about the concentration thing, I’ll look it up.</p>
<p>@Nerdlings: PR is usually very close to the actual test.</p>
<p>Hey I was just wondering if anybody had a copy of their 2008 exam with them…b/c i have a few questions about graphs that I won’t be able to type on here…</p>
<p>I’d really apreciate it if you could help me. I’ll try to help you too if i can. thanks!</p>
<p>“If i decrease concentration of Ni what happens to E and why?”</p>
<p>I’m pretty sure nothing since it’s not dependent on concentration.</p>
<p>sorry for getting back to you so late…
but the answers to the 4 questions i posted:
- b
- d
- b
- b</p>
<p>@tmanneopen</p>
<p>Thank you! Now I can go to sleep tonight (after cramming in as much electrochem as possible). :D</p>
<p>Well I literally got all 75 right on the second one and i missed two on the first, so I have a feeling they’re a lot easier than the exam…</p>
<p>FRQ is so-so on princeton review</p>
<p>Well, it shall be interesting to see tomorrow, haha. </p>
<p>Also, about the concentration thing. I’m sure that it does affect it, because a cell is an equilibrium.</p>
<p>I found out, it would decrease the voltage and increase the amount of zinc. The reason is that voltage is produced due to a cell trying to reach equilibrium. Since the cathode is naturally adding ions to the plate and the anode is naturally adding ions to the solution, the reverse would happen in this situation because the nickel is the cathode.</p>
<p>E goes down because the electric potential of the cell decreases. It is at equilibrium, and according to Le Chatelier’s Principle, the reaction will work in favor of the reactants. Therefore, the electric potential will decrease.</p>
<p>the equation for a cell not in equilibrium is:</p>
<p>E = E^o - (.0592/n)log(Q)</p>
<p>if Q increases, meaning reaction shifts right to favor products since Ni is being taken away, then E decreases</p>
<p>dang, equilibrium…anyone have a quick tip? I am completely blanking on it…
I’m looking at 2006 FRQ #1… and thinking frqqqqq.</p>
<p>any ideas how to do this?</p>
<p>Calculate the pH of the solution resulting from the adition of 20.0 mL of .100 M NaOH to 30.0 mL of 0.100 M HNO3.
a. 1.35
b. 1.7
c. 1.95
d. 2.52
e. 2.8</p>
<p>Calculate the pH of a solution prepared by mixing 300 mL of 0.10 M HF and 200 mL of 0.10 M KOH.
a. 2.82
b. 2.96
c. 3.32
d. 3.44
e. 3.53</p>
<p>And how would you write the equation for this?
Solid potassium oxide is added to water.
I thought it would be K2O + H2O yields KOH and H2O but it apparently yields 2K+ and 2OH- (how do you know whether the products are charged or not?)</p>
<p>Sorry I really don’t wanna look back in any posts lol but can anyone tell me a chemistry curve or calculating scores…I need a 4 really bad haha
What should I study the most to get that? lol</p>
<p>@sporty:
- i don’t think this would be asked since you don’t have a calculator during the MC…
but basically, you would find the concentration of HNO3, which is a strong acid.
so once you find that, just do pH = -log[HNO3] </p>
<p>and for the equation part…
since KOH is a a strong base, it would be written as just OH- and K+ since it disassociates completely.<br>
and you can just disregard the water…
basically though, you’re right.</p>