<p>I have an idea. What if we post practice questions for the AP Chem exam everyday, and whomever answers has to post another question so the cycle continues? Does anyone think this would be a good way to practice? </p>
<p>Btw: I'm not entirely new, I used to be registered as EarlySenioritis. </p>
<p>1st question:</p>
<p>Which change increases the mean free path of molecules in a sample of gas?
A. increase in pressure at constant temperature
B. increase in density at constant temperature
C. increase in temperature at constant pressure
D. increase in temperature at constant volume
E. increase in pressure at constant volume</p>
<p>OK, new question. Because I can't figure out the answer to the first</p>
<p>The electron configuration of the oxide ion is most similar to the electron configuration of the
A. Sulfide ion
B. Nitride ion
C. Oxygen atom
D. Nitrogen atom
E. Fluorine atom
my guess is "C" off the top of my head. Anybody else </p>
<p>I like B - then they both the electronic configuration 1s2 2s2 2p6.</p>
<p>8.One atomic mass unit (amu) is closest to the mass of
A. One proton
B. One electron
C. The oxygen-16 nucleus
D. The carbon-12 nucleus
E. 6.02 * 1023 atoms of hydrogen</p>
<ol>
<li>A, One Proton
The value of Plank’s constant is 6.63 * 10-34 J * s. The velocity of light is 3.0 * 108 m * s-1. Which value is closest to the wave length in nanometers of a quantum of light with frequency of 8 * 1015 sec-1?
A. 3 * 107
B. 4 * 101
C. 5 * 10-18
D. 2 * 10-25
E. 1 * 10-25</li>
</ol>
<p>Not interested in the round-robin, but since no one answered the first question, I thought I'd let you know it's C. For the mean free path to increase, the gas particles must be spaced farther apart. The only way for temperature to increase while pressure stays the same (as in C) is if volume increases, thereby increasing the mean free path.</p>
<p>I'll say A, You probably rounded in a diff. place </p>
<p>In the complete combustion of butane, the products are
A. water and carbon
B. butanone and water
C. butanoic acid and water
D. carbon dioxide and water
E. carbon dioxide and hydrogen</p>
<p>RESmonkey - your answer is almost right. The answer you got is in meters (since c is in meters). The questions asks for the wavelength in nanometers. Multiply your answer by 10^9 to convert meters to nanometers and you get B.</p>
<p>D.. kind of a giveaway. Trigonal pyramidal instead of octahedral would make it a trickier question, I think.</p>
<p>FightSenioritis' first problem - I think its E. The F atom has the exact same electron config as the O- ion, which is just the oxygen atom config with an extra electron. Agree?</p>
<p>
[quote]
The electron configuration of the oxide ion is most similar to the electron configuration of the
A. Sulfide ion
B. Nitride ion
C. Oxygen atom
D. Nitrogen atom
E. Fluorine atom
my guess is "C" off the top of my head. Anybody else
[/quote]
For the record, the answer to this is B. Look at the periodic table. A is a possibility (both oxide and sulfide have -2 oxidation state), but nitride and oxide have the same configuration. Anyway, good luck to all you AP Chem'ers.</p>