2010 AP Chemistry Form B Discussion

Test Preparation AP Test Preparation
1

<p><a href="http://apcentral.collegeboard.com/apc/public/repository/ap10_frq_chemistry_formb.pdf%5B/url%5D"&gt;http://apcentral.collegeboard.com/apc/public/repository/ap10_frq_chemistry_formb.pdf&lt;/a&gt;&lt;/p>

<p>Form A is out. Thanks Takayu.</p>

<p>Anybody took Form B?? </p>

<p>
[quote]
Just ask if you need an explanation/solution to one of these. One of my parents is an AP chem teacher, so these answers should be fairly accurate.</p>

<p>1A Ksp= [Ag+][Br-]
1B 7.1x10^-7 M [Ag+]
1C 3.55x10^-7 M [Ag+] which is less than the concentration in part B
1D V= 37618 L = 38000 L
1E Q=1.04x10^-3
Q>Ksp, so precipitate will form
1F Ksp for AgBr is greater than AgI since AgI precipitated to a greater extent than AgBr.</p>

<p>2A -3.2 degrees C
2B Endothermic - Temperature decreased, so energy was absorbed by the the system from
the surroundings. Endothermic solutions feel colder because they absorb energy while
exothermic solutions feel hotter because they "give off" energy.
2C (i) q= 1304 J = 1300 J
(ii) 15222 J/mol = 15 kJ/mol
2D 0.0701 kJ/mol K = 70. J/mol K
2E 12.46 kJ/mol = 12 kJ/mol
2F ΔH decreases. A Decrease in the temperature will cause a larger negative temperature
change for the solution. So for q=mCΔT, the ΔT would be larger and magnitude and
more negative, so since q basically equals ΔH, ΔH decreases.</p>

<p>3A (i) 0.00625 mol Cl-
(ii) V = 0.0797 L
3B (i) Cl- is 2nd order
(ii) MnO4- is 1st order
3C (i) r = k[Cl-]^2[Mno4-][H+]^3
(ii) 1.93x10^-3 M^-5 s^-1
3D Likely? I don't know the exact explanation to this one so....</p>

<p>4A OH- + H+ → H2O
Solution turns yellow since there would be more acid in the solution
4B C3H8 + 5O2 → 3CO2 + 4H2O
The solution will be acidic since CO2 will dissolve in water to form carbonic acid H2CO3
4C 2H2O2 → 2H2O + O2
O is has a 1- oxidation number in hydrogen peroxide</p>

<p>5A H-C≡C-H
5B Ethyne contains a ≡ bond so it has the shortest bond
5C (i) trigonal planar
(ii) tetrahedral
5D False- boiling disrupts the intermolecular forces, not the intramolecular bonds or
covalent bonds that hold atoms together.
5E Ethane or Ethyne are both polar
5F Ethanol is capable of H-bonding with water, where as Ethaneiol has only dipole-dipole
forces which are weaker than H-bonding</p>

<p>6A 1s² 2s² 2p6 3s² 3p6 3d10
6B Zn 2+ It is more difficult to remove electrons from a positive ion than from a neutral
atom.
6C Al(s) is oxidized
6D NO3- goes to the Al(NO3)3 solution to replenish electrons moving out of solution. K+
goes to Zn(NO3)2 compartment to replenish Zn2+ plating out on the Zn electrode
6E E = 0.90 V
6F E is positive, so the reaction is spontaneous. Therefore ΔG is negative for a
spontaneous reaction.
6G The cell voltage increases. Q= [Al3+] / [Zn2+] = 0.01 / 1.0 so Q is less than 1.
According to the equation Nernst equation, Ecell = E° - (0.0592/n)logQ , a Q less than
1 will cause log to be negative, and causes the rightmost term to become positive,
which will add to the value of E°

[/quote]
</p>

2

<p>Wow form A was so much easier than Form B… not fair T_T…haha</p>

3

<p>shoot for 5D i wrote that it was true, but mentioned intermolecular forces… do they give partial credit?</p>

4

<p>I took Form B… FRQ 1 was really unexpected! and I thought that redox was really over-emphasized!</p>

5

<p>I took form A. Part A was quite difficult, while part B was very easy.</p>

6

<p>

</p>

<p>I wrote unlikely because there are three molecules and it is improbable that they all collide to form a desired orientation.</p>

7

<p>5E should read “Ethane and Ethyne are both NONpolar”</p>

8

<p>3D Unlikely! If the reaction occurred in one step the rate law would be “rate=k[Cl-]^4[MnO4-][H+]^8”. And it’s not. It is also unlikely because the rate law would indicate that 6 reactant particles would have to collide at once.</p>

9

<p>Form B…
1A n-butane=saturated hydrocarbon C-C-C-C (surrounded completely by carbons)
isobutane=propane with a methyl group attached to the middle carbon
1B n-butane has a greater surface area b/c of its cylindrical shape; molecules have greater contact so its dispersion forces are stronger than those of isobutane; thus, n-butane has a higher boiling point
1C Kc=[isobutane]/[n-butane]
1D P= 0.24 atm
1E (i) Pressure remains constant. Moles of gas do NOT change.
(ii) [isobutane]= 0.0071 M [n-butane]= 0.0029 M
(iii) volume halves, concentration doubles [n-butane]= 0.0058 M
1F same as 1E (ii)???</p>

<p>2A cathode=reaction with Mn+2 and MnO4-
2B Ecell= 0.72 V
2C 5.0 mole e-
2D K=6.5 x 10^60
2E ???
2F ???</p>

<p>3A TeO2
3B +6
3C 3, 1, 8, 3, 2, 1
3D 1.81 x 10^-4
3E 1.25 x 10^-3
3F 0.598 g TeO2</p>

<p>4A CuSO4*5H20 → CuSO4 + 5H20 90.0g H20
4B Ni+2 + 4NH3 → [Ni(NH3)4]+2 Ni+2 acts as the Lewis Acid
4C CH3NH2 + H+ → CH3NH3+ basic solution</p>

<p>5A HCl starts with acidic pH
5B equivalence point= pH 7
5C .0040 mol titrant
5D Methyl Red pH range includes the equivalence point
5E no difference (TRICK QUESTION)
5F starts at pH of 13 and ends at pH of 1</p>

<p>6A first order
6B first order
6C rate=k[H2][Cl2]
6D M-1 s-1
6E 5.64 x 10-6 M-1 s-1
6F 2NO2 → 2N2 + O2
6G intermediate; it is produced and consumed during the reaction
6H step 1 is the slow step b/c it corresponds to the rate law</p>

<p>Hit me up with questions/comments!</p>

10

<p>Thank you for posting these answers tr0mb0n3man, they are really helpful. I was wondering if anyone had any idea on how to solve 2E and 2F because they confused me as well.</p>