<p>I did</p>
<p>KClO3 -> KCl + O2</p>
<p>C6H14 + O2 -> water + carbon dioxide</p>
<p>H+ + OH- -> water (dont remember the acid and the base)</p>
<p>Ni(OH)2</p>
<p>Zn+Cu+2 one</p>
<p>I am pretty sure.</p>
<p>is it ok to write only the flame color, yellow which i put, of sodium carbonate and conclude that the "other one" would be automatically calcium carbonate? I wasnt sure it was red or blue, so i vaguely ended it like that hahahahaha</p>
<p>and the hybridization ones.. they were sp3 and d2sp, triagonal pyramidal and rectangular pyramidal right? lol haha obsessing.</p>
<p>"H+ + OH- -> water (dont remember the acid and the base)"</p>
<p>Was it the ethanoic (acetic) acid one?</p>
<p>that's what I put at first, but then I remembered that acetic acid is weak...so I changed it to
HC2H3O2 + OH- --> H20 + C2H3O2-</p>
<p>the descriptive chem qustions were soooo easy--seriously--took me like 2 minutes to pick 5</p>
<p>choices 2 and 7 were easier for me; probably did bad though</p>
<p>"KClO3 -> KCl + O2"</p>
<p>Didn't it say potassium chlorate (ClO4)?</p>
<p>I think I missed the first part of #8.</p>
<p>oops..... oh well.</p>
<p>net-ionics were quite easy.....</p>
<p>thesea re the ones I did.</p>
<p>C6H14 + O2 --> CO2 + H2O
Ni+2 + OH- --> Ni(OH)2
Zn + Cu+2 --> Zn+2 + Cu
HC2H3O2 + OH- --> H2O + C2H3O2-</p>
<p>forget my last one.</p>
<p>i'm an idiot, lol. i said c2h6 was hexane. d'oh... i got the products right though (naturally, just forgot HEXANE) i still get 2/3 right? :P</p>
<p>yea..... my chem teacher said ALWAYS do the combustion reaction first cause you're guaranteed 2/3 points.</p>
<p>i wish they would've had a complex ion net-ionic........ I like those :(</p>
<p>why is it
HC2H3O2 + OH- --> H2O + C2H3O2-
and not just H+ + OH- --> H20
It says assume all reactants are aqueous unless otherwise noted. therefore C2H3O2- is a spectator ion and gets cancelled</p>
<p>Precocious: you should get 2/3 b/c products are worth 2 points and reactants are worth one</p>
<p>bigal, acetic acid is a weak acid...therefore it doesn't dissociate as H+ and C2H3O2 it stays as HC2H3O2...</p>
<p><em>Ahem</em></p>
<p>1.
a) i. Ksp = [I-]^2[Pb2+]
ii. 2.6E-3
iii. 8.8E-9
b) [Pb2+] = 1.3E-3 [I-] = 2.6E-3 (A saturated solution will always have the same MOLAR CONCENTRATION in solution. Doubling the volume will double the AMOUNT of moles in solution, but not the MOLAR CONCENTRATION.)
c) Decrease; cite the common ion effect or LeChatelier's principle
d) i. 4.1E-6
ii. Q = 1.7E-11 < Ksp (1.2E-10); since Q < Ksp, no ppt forms</p>
<p>3.
a. i. .6112 g C
ii. .3564 g N
iii. .2035 g O
iv. C4H5N2O
b. i. 188 g/mol
ii. C2H4Br2</p>
<ol>
<li><p>a. Compound X = Na2CO3; Compound Y = KCl; Compound Z = MgSO4
b. Mg(OH)2
c. Rxn: CO3-2 + H2O --> H2CO3 + OH-; increase in OH- concentration pushes pH above 7
d. Measure 33mL of stock soln with 50mL buret. Add 33mL stock soln to 100mL volumetric flask. Add distilled water until volume of soln is about 98mL. Use dropper to add distilled water until meniscus reads 100mL.
e. Lots of answers to this one. The SIMPLEST: Add both solids to water and stir. The one that dissolves is Na2CO3, the one that does not dissolve is CaCO3. </p></li>
<li><p>a. i. H-bonding, dipole-dipole ii. LDF
b. Glucose is polar and thus soluble in water (likes dissolve likes, polar molecules dissolve in polar liquids like water. Cycolhexane is nonpolar and thus does not dissolve in polar liquids such as water.
c. i. Process 1: Intermolecular H-bonds must be overcome. Process 2: H-O bonds must be broken.
ii. Disagree; when H2O boils, the intermolecular H-bonds must be overcome, but the H-O bonds remain intact.
d. i. Diagram 2: uncatalyzed reaction; Diagram 1: catalyzed reaction
ii. Disagree; adding a catalyst does not add any energy to the system, it simply provides an alternate path for reaction with a lower activation energy than the normal reaction. </p></li>
<li><p>a. 8s1
b. Q would be a metal, as it would lie at the bottom left of the periodic table. Additionally, all elements with their outermost electron in the S orbital are metals.
c. Q would have the largest atomic radius in its group because the periodic trend of atomic radius is decreasing as you move from left to right; this is supported by the fact that the Q atom would have an immense amount of shielding which would push valence electrons farther away from the nucleus.
d. +1
e. Q + H2O --> H2(g) + QOH
f. i. Q2CO3
ii. The compound would be soluble in water because all IA cations are always soluble in water.</p></li>
</ol>
<p>Bam.</p>
<p>you sure all the answers for #1 are correct?</p>
<p>im almost positive</p>
<p>zspot i like your answers...they seem like the ones I got too (at least the ones I remember)...i chose to do 2 and 7 though</p>
<p>all the answers look good....I don't think i missed more than 10 pts on FR...</p>
<p>what bout #7 and #4?</p>
<p>is a flame test ok for the last problem in the lab question?</p>
<p>Flame test is fine...as long as you put the right colors for the flames ;)</p>
<p>Awesome, I got many of the same answers.</p>