<p>varience - the isotope 62.93 of copper is more abundant than the 64.93 isotope of copper. If they were equal in percentage, then the AMU of copper (with both isotopes existant) would be 63.93, but it's not it's really 63.55</p>
<p>To get that, you must have more of the 62.93 isotope.</p>
<p>I HAVE ANSWERED QUESTIONS 3, 4, 5, 6, AND 7!</p>
<p>3
a) (i) 1st order with respect to I-
(ii) 1st order ith rspect to ClO-</p>
<p>b) (i) rate=k(I-)(ClO-)
(ii) k= about 611.76 L/(mol)(seconds)</p>
<p>c) (i) natural log of (H2O2)
(ii) units for k is equal to 1/(minutes)
(iii) elongate the line, MAKE SURE IT REACHES THE SAME Y VALUE AS THE GIVEN LINE</p>
<p>4
a) Zn + Ni(2+) ---> Zn(2+) + Ni
c) C2H2 - O2 ---> CO2 + H2O
f) BH3 + NH3 ---> BH3NH3
h) Pb(2+) + I(-) ---> PbI2
FORGOT MY OTHER CHOICE</p>
<p>5
a) Nitrogen condenses, Oxygen drives the flame even more and burns, Hydrogen violently reacts</p>
<p>b) CaO is acidic because metallic oxides form acidic solutions
SiO2 is apparently neutral because it is a tranition metal and will form sand
CO2 is basic because nonmetallic oxides will form basic solutions</p>
<p>c) (i) Ag2S
(ii) KNO3 forms an aqueous solution and AgCl forms a precipitate that isnt black
(iii) Solution 1=AgNO3
Solution 2=Na2S
Solution 3=Kcl</p>
<p>6
a) If you dont know these, I hope you fail this exam. However, CF4 creates a tetrahedral shape, PF5 has a trigonal bipyramid shape, and SF4 has a seesaw shape.
b) (i) The bond angle is 109.5 degrees
(ii) The hybridization is sp3d (or dsp3). Either one works.
(iii) Like stated above, a seesaw or distorted tetrahedron.</p>
<p>c) (i) sigma= 4
pi= 1
(ii) I didnt know what formal charge was, so I just said structure one because phosphorus has 5 valence electrons and the oxygen double bonded to it and the 3 flourines bonded with all five. Structure 2 bonded with only 4 of the 5 valence electrons.</p>
<p>7
a) (i) NH3 enters into hydrogen bonding and NF3 contains London dispersion forces.
(ii) Hydrogen bonds are very strong compared to London dispersion forces, thus NH3 has a higher boiling point.</p>
<p>b) (i) KCl and NaCl both contain London dispersion forces but NaCl forms as a crystalline lattice. This was pretty darn tricky.
(ii) It takes more energy to melt a crystal rather than molecules bonded together with London dispersion forces. So NaCl has a higher melting point.</p>
<p>c) (i) All are at the 3rd energy level, so n=3.
(ii) Chlorine has the most valence electrons an the smallest radius, so it has the greatest pull on electrons, so it has the highest ionization energy. Silicon has the least amount of valence electrons and the largest radius, thus having the weakest pull on the electron, so it has the least ionization energy. Phosphorus is in the middle.</p>
<p>d) I DONT SEE WHY THERE IS MUCH DEBATE OVER THIS QUESTION. IT IS VERY STRAIGHTFORWARD.
(i) The average of the 2 masses is 63.93, so look at the periodic table and ta-da! Copper s mass is 63.55. Copper is the final answer.
(ii) OK, this is where some dont understand the question. I will show you in the form of an equation:</p>
<pre><code>Okay, you have the 2 isotopes, right? So, there is a combination of those two that equates to the mass on the periodic table. So, one of them composes x percent of the CU mass you see on the periodic table, and the other composes 1-x percent. So, you can come up with this equation: 62.93(x) + 64.93(1-x) = 63.55
I hope you can do the math, and once you do, you find that x equals .69
So, you can find 69% of the first isotope, and 31% of the second. thus, the first isotope is more abundant.
BUT, if you can understand this question, then you can see that the periodic table mass of copper (63.55) is less than the average mass of the 2 isotopes, so there must be more of the first, less weighing isotope.
</code></pre>
<p>IF YOU THINK I SCREWED UP, PLEASE REPLY!!!!!</p>
<p>Yea, you did screw up.</p>
<p>First, how can you make the line reach the same value? The graph is not big enough for that. As long as the line is above the original, that's correct, cuz less peroxide is used up without the catalyst..</p>
<p>Secondly, no, metallic oxides don't produce acidic solution, they produce basic solution. And vice versa, non metallic oxides (CO2) form acidic solutions.</p>
<p>Thirdly, NaCl and KCl ARE NOT londons. They are ionic. And also, NF3 is not london, it's dipole dipole. "I CAN'T BELIEVE YOU DIDN'T KNOW THAT. I HOPE YOU FAIL THE EXAM."</p>
<p>Gosh, some people on this board are such elitist JERKS. Geez.</p>
<p>Dude, that is not the purpose of a catalyst. A catalyst will use up the same amount of H2O2, but at a faster rate. And just because the line wasn't long enough, you could have still drawn a long line. So, I do think that the line should represent the same amount of H2O2 used up.</p>
<p>Yes, I messed up the acidic and basic thing, but ONLY ON THIS SITE. Thank god I didn't o that on the real thing.</p>
<p>And okay, fine, I guess I was stupid for the lonon ispersion thing. And I am sorry if I offended anyon by the xam failure comment. But seriously, if you didnt know anything about molecular geometry, I don;t know what else to say.</p>
<p>
[quote]
But seriously, if you didnt know anything about molecular geometry, I don;t know what else to say.
[/quote]
</p>
<p>But, seriously, if you don't know anything about being a courteous and nice person, then I don't know what else to say...</p>
<p>Holy crap, gt over it dude!!! I said I am sorry, okay? And plus, everyone would have gottn it right anyway. Dear lord.</p>
<p>yes, the same amount is used up, but not as fast. SO, on the amount of space they gave you, the line should end higher, because less is used up. You are right, eventually the same amount will be used up, because none be left. But that graph does not show the whole reaction, actually, we don't really know what it shows. Also, the Ln(O) is undefined. So we def don;t know when it's all used up (and the table doesnt show all the data either). </p>
<p>So, the line should just be ''higher'', or with a slope that's not as steep.</p>
<p>actually, that isn't true, because of the fact that they give you a table of concentration adn time values. If you look, you can see that the end of the graph is at (10, .61).</p>
<p>AND, I think you believe that we can't go farther than the X-axis's limits. But where does it sy that in the problem. If we were to stop at that particular point, then yes it would be above, but it never said that you can't continue off and longate the x-axis. Sinc it never did, one can elongate it.</p>
<p>I am right, and so are you. I just elongated the X axis and finished the graph. You are right in the sense that you only used whaat was given and stopped there.</p>
<p>I only feel that I am more right because one would think that you should show the whole usage of the reactant, ven if it means elongation of the x-axis. Otherwise, like I said, we are both right.</p>
<p>email me at <a href="mailto:brsrkdb8er.4life@gmail.com">brsrkdb8er.4life@gmail.com</a> if you want a pictorial explanation. I made one, and we can come to a definite conclusion.</p>
<p>ok, i see what your saying. both answers are right (they're the same thing, you just completed it more). cool cool.</p>
<p>I am looking for the weighting of free response questions on the A.P chemistry exam for 2004 and the composite scores</p>
<p>Just took the AP, and it wasn't as bad as i thought it would be. I think i actually did better on free-response than MC. Can anyone tell me how to calculate raw score? I tried using the 2002 Scoring Sheet on the Collegeboard website, and I got in the range of 108-111. Is this high enough to get a 5? Is the scoring the same every year? Thanx</p>
<p>What's all this flame stuff I'm reading? I don't recall any flame questions from the free response. Was there more than one version of the test? Maybe you guys are talking about 7 or 3, because those are the two that I didn't do.</p>
<p>MC: it was okay, slightly easy, I think I got 80% of them right, i left 6 blank.</p>
<p>Free Resposne: Reactions were easy, #1, #3, #7 were all easy, i think I got 100% on that. #5 and #6 were hard, probably got half credit.</p>
<p>Any ideas on my score?</p>
<p>wow ok looks like some perople hjad different verion of the test</p>
<p>The FR was pretty good. The MC, though, will determine a three, four, or five. I'm praying for a five, but in most likelihood, its probably a four or even a three. Wahhh. I should've studied harder.</p>
<p>the flame answers you're most likely hearing about is how to determine the difference between CaCo3 and Na2Co3. When you put CaCo3 in a flame it turns red.</p>
<p>THIS is LAST year's thread. Please CREATE a NEW one or REPLY to another one ALREADY CREATED for THIS YEAR's TEST.</p>
<p>i just thought i was ****ed. afterall it wasn't this year's test....</p>
<p>go to the ap test subforum, there are like 3+ different ap chem topics, haha.</p>