FRQ for AP Chem. 2009 Debate

<p>This is what I did. Oh yeah, can the editor please delete my old thread because I didn't specify which FRQ it was pertaining to?</p>

<p>2.
a. (1.18)(.843)= 9.95 x 10^-1 g</p>

<p>b. 157.70-.995= 156.7 g</p>

<p>c. 158.08-156.7= 1.38g</p>

<p>d. Use PV=nRT or (PV)/T=(PV)/T, either are fine
n= 1.38/x (x is the molar mass)
(750/760)(.843)= (1.38/x)(.0821)(273+23)
x= 40.3 g</p>

<p>e. (44-40.3)/44 x 100= 8.41%</p>

<p>f.
OC 1:Yes, since dry air is less dense than the gas, the gas may appear lighter
OC 2: No, since gasses are more dense at cooler temperatures, that would make the gas heavier, which is not what happened</p>

<p>g. Measure 843 mL of water out and see if it overflows, fits perfectly, or doesn't fill the container. The container would be less than 843mL, exactly 843 mL, or more than 843 mL, respectively. </p>

<p>3.a. i. 25/16=1.56 mole; since we need 3.125 mol of Cl2 to react with 1.56 mol of methane, Cl2 is the LR.</p>

<p>ii. 2.58 mol of Cl2 reacts with 1.29 mol of CH2Cl2, we need 1.29 mol of CH2Cl2.</p>

<p>b. (242000)/(6.022x10^23)=4.02x10^-19</p>

<p>c. I'm not really sure about this one.
E=hv (v=c/lambda) so E=h(c/lambda)
4.02x10^-19=(Plank's)((3x10^8)/lambda)
4.95x10^-7 meters</p>

<p>d. Intermediate, since it is consumed.</p>

<p>e. Not sure about these.
i. 2
ii.1
4.a.*
i. 2(Fe2O3) + 3C = 4Fe +3(CO2)
ii. 0 to +4
b.
i. NH3+HCL=NH4Cl
ii.acidic, since NH4Cl can donate protons (not sure about that)
c.
i. 2(HgO)=2Hg+O2
ii. less than since gas escapes</p>

<p>5.a. [H2][CO]/[H2O]
b. decrease, since there are less gas molecules on the left side, which is where equilibrium will shift
c. less than one, since gibbs is positive, it is not spontaneous, and will therefore prefer the reactants
d. the products, since it is endothermic, and the products have gained energy compared to the reactants
e. yes, because the gibbs for y is closer to 0 than for x, and thus will occur more easily
f.
i. it's negative, since the products have less entropy than the reactants
ii. i forgot what i put, but it was negative (I think -172kJ/mol)
iii. increase, since it will react to turn into two moles of carbon monoxide, which has more volume than solid carbon and one mole of carbon dioxide
6.
a. i. basically 1s2 up to 3p4 (it's hard to type out) for S, and up to 3p6 for S2-
ii. since the same pull of the nucleus is now spread out over more electrons
iii. I put S2-, since it will be more strongly attracted to the positive end of the magnet</p>

<p>b. it is easier to remove from S2-, since the Ar atom has more protons and therefore a stronger pull</p>

<p>c. I put the p atoms. Not sure.</p>

<p>d.*
i. I put the H2S, but not sure.
ii. I put H2S, but I think H2O is stronger. I think I got d wrong.</p>

<p>I basically did what you did with the exception of the PV=nRT. I wasn’t sure on the d either. I put H2S for both as well.</p>

<p>Shouldn’t the Ammonia one be NH3 + H+ –> NH4+ since it’s a net ionic?</p>

<p>Yeah, but I did NH3 + H30+ => NH4+ + H2O</p>

<p>There isn’t a precipitate so I think it is what I put. Anybody know for sure?</p>

<p>ii.acidic, since NH4Cl can donate protons (not sure about that)</p>

<p>My explanation would be that NH3 is a weak base; NH4+ is the conjugate acid of NH3 and therefore a weak acid. What do y’all think?</p>

<p>It’s acidic, but it’s because NH3 is a weak base and HCl is a strong acid. It said you added equal volumes of both, so there would be extra H+ left over, even after the base was neutralized.</p>

<p>Oh, and:</p>

<p>2HgO => 2Hg + O2</p>

<p>2Fe2O3 => 4 Fe + 3O2</p>

<p>^Is that right?</p>

<p>I put Hg2 instead of 2Hg. I think you left out solid carbon in your reaction.</p>

<p>i put mass was conserved for the Hg reaction</p>

<p>For Mercury, I think it makes sense. Also, check this link,
[WikiAnswers</a> - What is the decomposition reaction for mercury oxide](<a href=“http://wiki.answers.com/Q/What_is_the_decomposition_reaction_for_mercury_oxide]WikiAnswers”>What is the product of the decomposition reactant of mercury oxide? - Answers)</p>

<p>And the Iron oxide reduction is in the link towards the bottom.
[WebElements</a> Periodic Table of the Elements | Iron | Essential information](<a href=“http://www.webelements.com/iron/]WebElements”>WebElements Periodic Table » Iron » the essentials)</p>

<p>Yes!!! Got the reaction with iron oxide right. I’m so bummed that I got the Hg reaction wrong. Is mass conserved though?</p>

<p>For 6d: </p>

<p>i. H2S has more significant London dispersion forces… I only said because S is larger and thus has more electrons, which makes its dipole moments more significant… but I think there are one or two other reasons.</p>

<p>Oh, just found this:</p>

<pre><code>* Larger and heavier atoms and molecules exhibit stronger dispersion forces than smaller and lighter ones.

  • In a larger atom or molecule, the valence electrons are, on average, farther from the nuclei than in a smaller atom or molecule. They are less tightly held and can more easily form temporary dipoles.
    </code></pre>

<p>([London</a> Dispersion Forces](<a href=“http://www.chem.purdue.edu/gchelp/liquids/disperse.html]London”>London Dispersion Forces))</p>

<p>ii. H2O has more dipole-dipole interactions b/c O is more electronegative… hydrogen bonding, the strongest type of dipole-dipole interaction, occurs.</p>

<p>(to Poseur) Hydrogen-bonding is just a strong, special type of permanentdipole-permanentdipole interactions</p>

<p>the london forces question was unfair IMO because they have little effect in polar molecules (which H20 and H2S are). Only in nonpolar do they even matter.</p>

<p>regular sulfur has unpaired electrons–>paramagnetism</p>

<p>idk what the thing at the top of my previous comment is</p>

<p>nvm its not there anymroe lol</p>

<p>I put 2, 0 for the kinetic order one. Not sure.</p>

<p>My rationale: 2 because it’s a fast/slow reaction. you add the reactant of the first reaction to the 2nd one to make the rate law equation. 0 for two because fast reactions don’t matter??</p>

<p>isn’t it lab value - expected value/expected value x 100 for percent error? so the answer is negative?</p>

<p>^well the numerator has absolute value bars surrounding it so it’s always positive</p>