Student

Question 5
a. Nitrogen: Splint goes out completely (extinguishes)
Oxygen: Splint reignites or glows brighter.
Hydrogen: Splint explodes/"pops"

b. CaO + H2O --> Ca(OH)2, a strong base. Hence, pH > 7.
Silicon oxide is a neutral compound. pH = 7.
CO2 + H2O --> H2CO3 --> H+ <-> HCO3-
HCO3- <-> H+ + CO3^2-
Since H2CO3 can ionize twice, its pH < 7.

c. Black ppt is Ag2S (s)
Mixing solution 1 with soln 3 yields the AgCl (s), a white ppt.
Solution 1: AgNO3 (Silver nitrate)
Solution 2: Na2S (Sodium sulfide)
Solution 3: KCl (Potassium chloride)

Question 6

a. CF4: C is central atom. 4 single bonds.
PF5: P is central atom; 5 single bonds.
SF4: S is central atom; 4 single bonds. S atom has 2 additional electrons.

b. Approx. 109.5º
PF5 has sp3d hybrid.
SF4 has a seesaw molecular geometry.

c. 4 sigma + 1 pi (from the double bond)
Structure 1 represents the molecule better because each atom has a formal charge of 0, hence making it more stable.

Student

Question 8

a. Change in G is negative because the dissolving of AgNO3 into its component ions in water is a spontaneous process. Any compound with the NO3 ion is always soluble in water.

b. Change in S is positive because a solid object's dissolution increases its disorder.

c. The sign of H is positive because the reaction is endothermic: heat must be absorbed.
Yes, it is consistent. However, TS needs to be more positive than H in order to yield a negative value for G.

d. Ox: Na (s) --> Na + e-
Red: O2 + 4H+ + 4e- --> 2H2O
Oxidation rxn takes place at the anode
G is always positive for electrolysis rxns because work must be applied.

Hope this helps, guys!

Student

Crypto86, I think you need to use the Henderson-Hasselbalch equation (for buffers) in c-ii. Adding sodium propanoate yields the NaC3H5O2-HC3H5O2 buffer (a weak acid and its salt). I may be wrong, though.

everything else in question 1 looks right to me.

Student

Question 4, continued
b. Al(OH)3 + KOH --> KAlO2 + 2H2O
f. BF3 (g) + NH3 (g) --> H3NBF3 (s)
g. SO3 + 2 NaOH --> Na2SO4 + H2O

These were indeed hard!

man_on_fire

Dude, why did you add the hydrogen concentration to your answer at the end of this part Crypto??? I am a 100% sure you're noit supposed to do that. My AP Chem teacher did this part of the problem on the board and he didn't add the hydrogen concentration(from part a) to the end of this part of the problem. Why would you do that??? Whatever, it doesn't really change the answer much.

bsbllallstr8

question 4...
CaCO3+HC2H3O2 ---> Ca(C2H3O2)2 + CO2 + H2O

Cant write H+ for a weak acid.. could prolly also be written Ca2+ + C2H3O2... but dont know for sure

Variance2004

if anyone wants to discuss the exam w/ me..
IM silvachik08 on AIM...

we can discuss it fast!!

crypto86

man on fire - remember, the original solution has [x] concentration of the C3H5O2- ion in it already. Remember - it's [x]^2 that equals 3.66*10^-6; the [H+] times [C3H5O2] = 3.55*10^-3, so [H+] = [C3H5O2-] = 1.88*10^-3M

Therefore, you need to consider the already existing C3H5O2- ion concentration already in solution; you can't just disregard it from part a. That's why I added 1.88*10^-3 (meant for C3H5O2- concentration, not H+) to the ion added from the NaC3H5O2, because some ion is already in solution before you add the solid. I'm pretty sure I'm right on that, but I'll double check again.

Also, student, for the solid Al(OH)3 and KOH solution reaction, I had:

Al(OH)3 + OH- --> [Al(OH)4]-

KOH is a very strong base (Kb > 1) therefore it is always ionized. So the first ionic equation would be:

Al(OH)3 + K+ + OH- --> [Al(OH)4]- + K+; and the K+ would cancel out on each side, leaving: Al(OH)3 + OH- --> [Al(OH)4]-

Hecatonchires

Just a little comment about the reaction questions. I personally thought they were the easiest of any reaction Q I've seen, but then again, my teacher did prepare us with weekly quizzes for a little. I mean, a electrochem rxn, a combustion rxn, and a few others were really simple.

I chose:

Zn + Ni+2 => Zn+2 + Ni (battery)
C2H2 + O2 => CO2 + H2O (combustion)
Li + N2 => Li3N (appeared on a previous test)
BF3 + NH3 => BF3NH3 (appeared on a previous test)
Pb+2 + I- => PbI2 (common lead iodide demo rxn)


And, with the previous answers given for equilibrium, I agree completely.

Hecatonchires

I agree except for the units in (c). The units would just be minutes^-1, since the rxn is first order, rate = k[H2O2].

Student

Corrections

3c. Units for k are 1/min.
4d. CaCO3 (s) + HC2H3O2 (aq) ---> Ca2+ (aq) (C2H3O2)- (aq) + CO2 (g) + H2O (l)

HELTAHIR

For that one reaction that was "Ammonia is added to Boron Trifluoride" and the answer was H3NBF3, how many points do you think they would take off if i just wrote HNBF3?

HiWei

Probably just one, I mean it wasnt that big of a mistake.

Variance2004

H3NBF3 destroys the ozone layer.. remember..

grumpybear727

for SO3 + NaOH, wasn't the product HSO4- ?