<p>shanianwang's first equation is wrong.</p>
<p>KClO3 --> O2 + KCl</p>
<p>looking back at it, what was I thinking?</p>
<p>On lab question #5, they ask you to write the formula for the precipitate. What if I wrote the whole reaction, including the precipitate? Do I not get any credit?</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>For 2:
a) One must reverse the top reaction to achieve the final: the answer that I got was -283 kJ/mol.
b) -86.55 J/mol-K
c) -367.7 kJ/mol
d) Yes, ΔG° was negative.
e) e^160.92, which I can't calculate now.</p>
<p>shanianwang, your NH3 equation is wrong. </p>
<p>NH3(g) + HF --> NH4+(aq) + F2(g)</p>
<p>are you sure that one is not:</p>
<p>NH3 + HF --> NH4+ + F-
(could you explain why perhaps?)</p>
<p>um not -367.7.... i got -257 </p>
<p>and got 1.47 X 10^45.</p>
<p>Fluorine is the best oxidizer in the business. It will stop at nothing to gain an electron and become isolectronic to Neon. F- will never stay by itself, ever. It's tooooo damn reactive. When two F- see each other, they latch on and share the electron like nobody's business and form F2. While technically the reaction does free up F-, it is present in solution for hardly a moment. I don't know if they'll accept F-, cuz it's never present in solution.</p>
<p>Gyros321, I recieved the same answer for K, so something is wrong. K requires a correct ΔG.</p>
<p>wow zpot nice reasoning...... I think I missed that net-ionic.</p>
<p>for one reaction i put</p>
<p>NH3 + H+ ---> NH4+</p>
<p>Two questions, why do you need the HF in the equation? is it because it is a weak acid?</p>
<p>Also, assuming this answer is wrong, i would get no points? or what?</p>
<p>Yea you would either get 0 or 1 because SOMETIMES (on the old tests) they give partial credit.</p>
<p>tim555, I think you have to get all the reactants to get 1 point, and then one of the products to get 1 of the 2 product points.</p>
<p>Regarding NH3 + HF:</p>
<p>from Princeton Review Cracking: HF + NH3 --> NH4+ + F-</p>
<p>Yeah, I've seen that acid-base rxn several times. </p>
<p>It's NH3 + HF --> NH4+ + F-</p>
<p>How would you even correctly balance NH3 + HF --> NH4+ + F2</p>
<p>you could go 2NH3 + 2HF --> 2NH4+ + F2, but that wouldnt be balanced because of charges. btw, I know you don't need to balance the reaction, but i'm just demonstrating why that can't be right.</p>
<p>tim, you need HF rather than simply F- because HF is a weak acid and doesn't dissociate completely.
for reactions, you get 1 point for the correct reactants and two points for correct products. You might get zero points for that answer. </p>
<p>oh and zspot, for number 8, are you sure that the reaction for Q was:
Q + H2O --> H2(g) + QOH.</p>
<p>I put Q + 2 H20 --> H30+ + Q+ + OH-
But that was the one I was least sure about also.</p>
<p>hey guys for # 7 were the two shapes trigonal planar and trigonal bipyramidal</p>
<p>the reaction for Q and water was....</p>
<p>Q + 2H2O --> H2 + Q+ + 2OH-</p>
<p>Princeton review says:</p>
<p>CH3COOH + OH- --> CH3COO- + HOH</p>
<p>I am 100% certain, our teacher told us so and our textbook confirmed it.</p>
<p>good thing i didn't do that one :).</p>
<p>can't wait to see what we all got...</p>
<p>Would I get a point if i did
H + oh --> h2o?</p>
<p>I mean I had h2o D:</p>