The AP Chemistry Study Thread

Test Preparation AP Test Preparation
101

<p>yea scareya that's right.
HSO4- goes to SO4-2 or H2SO4</p>

<p>btw, on sig figs, this is what I do.
forget about sig figs (wait, keep reading)
do all the calculations needed to find a specific answer (one that you'll box and submit as a final answer).
revert back to sig figs, see how many numbers to use, see how many to convert to 0.</p>

<p>also if your having major problems w/ sig figs, try this.</p>

<p>difference between 130 and 130.</p>

<p>write it in scientific notation - 130 = 1.3 x 10^2
130. = 1.30 x 10^2</p>

<p>realize it's not the best example.... but decimal points make EVERY number significant.
100000000000000 has one sig fig.
100000000000000. has 15 sig figs.
100000000000000.00 has 17 sig figs.</p>

102

<p>Scareya~ you are right.
Here is a question:
__ Cr +<strong>CrO42- +</strong>_ ---> ___ Cr (OH)2</p>

<p>If one mole of CrO4(-2) is oxidizes Cr according to the reactaion represented above, how many moles of Cr (OH) 2 can be formed?
A) 2
B) 3
C) 4
D) 5
E) 6</p>

<p>-2 is CrO4's charge</p>

103

<p>i'm assuming the blank is H+.
otherwise this turns into a redox........ (bad)</p>

<p>if its like that, then
__ Cr +<strong>CrO42- +</strong>H+ ---> ___ Cr (OH)2</p>

<p>my first step
__ Cr +<strong>CrO42- +</strong>H+ ---> <em>2</em> Cr (OH)2</p>

<p>Cr's balance, O's balance, H's don't.</p>

<p>__ Cr +_<em>CrO42- +</em>4<em>H+ ---> _2</em> Cr (OH)2</p>

<p>therefore the answer is A) 2.</p>

<p>gonna do it another way -- assume it's not H+, but just a blank, and do redox.</p>

<p>actually, it's the exact same thing, cept w/ 2e- on the left side.</p>

<p>so the answer is A...... let me know if i'm right.</p>

104

<p>Evilbooyaa~ you have to decide if it is redox or not. I got this from a sample AP Exam and the blank was blank. You can put whatever you think is right in that blank. Yes you are right. I am not giving anymore qs. Only person actually providing them today :(</p>

105

<p>My last question is just too hard to post up with numbers so here is another one...</p>

<p>Question: 2000 AP FR</p>

<p>.345 g BeC2O4 dissolved in H2O to produce 100. mL solution. 20 mL portion of solution titrated with KMnO4. Eqtn for reaction:</p>

<p>16H+ + 2MnO4- + 5C2O4---> 2 Mn 2+ + 10CO2 + 8H2O</p>

<p>Volume of .015 M (MOLARITY) KMnO4 required for equiv. point is 17.8 mL</p>

<p>a) Calc. # moles of MnO4- and C2O4 2- of each reacted at equiv. point.
b) Calc. # moles of C2O4 2- present in the 100. mL prepared solution.</p>

106

<p>new question:
design an experiment to create a buffer of pH 5, 100mL that will be capable of buffering 6.0*10^-5 mol of acid. you are given "unlimited" HC2H3O2 powder and "unlimited" .100M NaOH. (you need to detemine the amount of powder and solution you actually need for the buffer).</p>

107

<p>ahhhhhhh!!! you posted before me, i'll do yours later when i get more time, unless some1 else does before me.</p>

108

<p>tryingmybest~ at equivalence point, moles C2O4 should equal the moles of MnO4
MaVa=MbVb. Also M x L = moles
.015 M) (.0178 L)= 2.67 x 10^-4 moles of MnO4= 2.67 x 10^-4 moles C2O4 </p>

<p>so if there are 2.67 x 10^-4 moles C2O4 in .020 L, there are X in .1 L
X= 1.34 x 10 ^-3 moles in .100 L. </p>

<p>Is it me, or is there too much extra information. Well since this question came from an AP, I will guess that there were more question on this problem.</p>

109

<p>Here's an easy question for you guys:</p>

<p>Indicate whether the following balanced equations involve oxidation-reduction. If they do, identify the elements that undergo changes in oxidation number and the oxidation numbers.</p>

<p>a. PBr3(l) + 3H2O(l) -> H3PO3(aq) + 3HBr (aq)
b. NaI(aq) + 3HOCl(aq) -> NaIO3(aq) + 3HCl(aq)
c. 3SO2(g) + 2HNO3(aq) + 2H2O(l) -> 3H2SO4(aq) + 2NO(g)
d. 2H2SO4(aq) + 2NaBr(s) -> Br2(l) + S)2(g) + Na2SO4(aq) + 2H2O(l)</p>

110

<p>Does anybody know of a site where I can find info on the color of different types of acid/base indicators? (ie penophthalein sp?)</p>

<p>sry scareya, i'll get back to you q in a sec.</p>

111

<p>Is this what you wanted?
<a href="http://www.101science.com/PH.html%5B/url%5D"&gt;http://www.101science.com/PH.html&lt;/a>
<a href="http://en.wikipedia.org/wiki/PH_indicator%5B/url%5D"&gt;http://en.wikipedia.org/wiki/PH_indicator&lt;/a&gt;&lt;/p>

112

<p>stanfordream- HC2H3O2 is actually a solution (like NaOH), but i'll go with a powder I suppose..... maybe it is. never seen HC2H3O2 powder.</p>

<p>wait a second.
how do you make a buffer out of HC2H3O2 and NaOH?
you need NaC2H3O2 or something of the sort...... you need the acid and the anion/cation to produce a buffer..... you can't make it out of HC2H3O2 and NaOH...... can you?</p>

<p>help please!</p>

<p>Scareya- ill go thorugh the first one step by step
A)left- P=+3, Br=-1, O=-2 H=+1
right- H=+1, O=-2, P=+3, Br=-1</p>

<p>therefore, don't involve oxidation/reduction</p>

<p>2)'I' goes from a -1 to +5--> oxidation
'Cl' goes from +1(?) to -1--> reduction</p>

<p>3)'S' goes from +4 to +6--> oxidation
'N' goes from +5 to +2 --> reduction</p>

<p>4)'Br' goes from -1 to 0 --> oxidation
'S' goes from +6 to +4(in the S)2, which I assume to be SO2)--> reduction</p>

<p>stanfordream's question is still on the board, so I won't post a new one.</p>

<p>btw, sristi, don't feel bad about posting questions! it helps those too lazy to look for questions themselves</p>

113

<p>hint: to get NaC2H3O2(aq), HC2H3O2 + NaOH --> NaC2H3O2 + H2O</p>

114

<p>ah...... true.</p>

<p>meh someone else will do it. I gotta go do other stuff real quick.</p>

115

<p>Ah thank you scareya.</p>

116

<p>You're welcome.</p>

117

<p>What's the difference between a galvanic cell and an electrovoltaic cell?</p>

118

<p>Aren't they the same thing?</p>

119

<p>galvanic cell and a voltaic cell are the same, I think. not 100% sure. I never heard of electrovoltaic. I think you are mixing it up electrolytic cells. voltaic cells produce electricity. electrolytic produces electricity.
Edit: I meant electrolytic requires electricity. Sorry!</p>

120

<p>Srisiti, you are 100% correct. Galvanic cell = voltaic cell
It says that in my textbook.</p>