<p>@flamingmango i did the same thing for f & ii, my sig figs are just one off which is okay because they allow it :D</p>
<p>how many points do i get off for sig figs?</p>
<p>OHH ok nvm they allow one off which is good :P</p>
<p>@FlamingMango, I’m pretty sure you were supposed to use 1.86 in your calculation for the current, because rounded answers are not to be used in further calculations according to my teacher. idk though</p>
<p>That sigfigs bit isn’t something they would take off points for, I don’t think, since you could use a more accurate value of Faraday’s Constant (I used 96485) And get a slightly different answer.</p>
<p>On the sheet of constants however, they put the constant as 96,500.</p>
<p>I don’t believe they take points off for sig figs. These answers are graded more wholistically… As long as you didn’t go out to like 15 decimal places I’m sure you’ll be fine. and yeah i used 96,485 for farraday’s constant</p>
<p>For 3 did you guys get…
a) -571.6 KJ
b) -1418 KJ
c) 88 KJ - now that I think about this it may be wrong idk that’s what I put tho
d, e, f) already went over
g) ??</p>
<p>You can look in previous years, when they ask you to calculate moles (or whatver) in one part, you are to use that same value in the next part with your next calculation.</p>
<p>But like someone said, you get ±1 sig fig tolerance, and i believe you can only get 1 point max off per problem for sig figs.</p>
<p>I got a different answer for b, how did you do it?</p>
<p>Also, how did you guys do 2. c)?</p>
<p>3c basically heat of vap for h20 is 44.0 kj/mol. thus H20 (l) –> H20 (g) is 44.0 per mol. Let x = heat of formation for h20 (g). (x) - (2(-285.8)) = 44. </p>
<p>and then you multiply x by two, and that is the answer.</p>
<p>anyone care to agree or disagree?</p>
<p>10g H2 x 1molH2/2.016gH2 x 1molH2O/1molH2 gives you 4.96mol H2O. then you multiply that by -285.8KJ. anyone else get something different? how did you do yours</p>
<p>so firered you’re basically saying the answer is 88? I followed what you did but on the test all i did was 2(44) i guess i completely missed the point. nevermind yeah i see what youre saying wow i totally missed that. 3c is NOT 88.</p>
<p>@Pinkrose Yes, but they certainly don’t take off for more exact values! :)</p>
<p>I did it like 10 g <em>(1 mol H2/2g)</em>(2 mol H20/2 mol H2)*(285.8kJ/2 mol h20)=714.5</p>
<p>you did the same as me except you put 285.8 over 2 mol instead of 1. doesnt it tell you in the problem 285.8 is per mol of h20 already??</p>
<p>#3. (c) Using Hess’s law:</p>
<p>2H2(g) + O2(g) –> 2H2O(l) dH = (2)(-285.8) = -571.6</p>
<p>[ H2O(l) –> H2O(g) dH = 44 ] X 2 , multiple by 2 so equations cancel</p>
<p>2H2(g) + O2(g) + 2H2O(l) –> 2H2O(l) + 2H2O(g) dH = -571.6 + (2 X 44)</p>
<p>Simplifies to 2H2(g) + O2(g) –> 2H2O(g) dH = -483.6 ≈ **-484. kJ/mol **</p>
<p>That’s what I did. A lot of people i talked to said they just did 2 X 44 = 88 but that doesnt make much sense to me. I dunno whats right, but thats how i did it</p>
<p>well, i got that one wrong too i guess, -.-</p>
<p>Yeah flamingmango that has to be right. I just looked at the problem and put 88, but my answer is definitely wrong. that’s the right method. what did you put for part g?</p>
<p>@flamingmango, i did c the way you did it too</p>
<p>Anyone care to explain how to do #1 c & d?</p>
<p>Part A</p>
<p>1.) Don’t even remember what I did but I think I did it correctly.</p>
<p>2.) i) was pretty straight foward
ii) for lab procedures I did
- put on goggles/gloves
- pour x amount of stock into beaker
- pour the beaker into the cylinder
4.) started #3 and came back when I had 2 mins left I wrote “blow up the lab” :O</p>
<p>iii) something along the lines of not needing to be very accurate</p>
<p>iv) i put nacl because i was thinking of the neutralization of hcl im wrong though :/</p>
<p>b) no idea, didnt have time to think about the problem
c) i got 1% hahaha didn’t have time when i came back to finish it</p>
<p>3) 1.86 mole of e-?, and 300 something amps, i didn’t do correct sig figs = fail</p>
<p>part b:
4) a. i got this one
b. fail didn’t know what coordination complex was
c.) i said ppt will form but i didnt say which one :S</p>
<p>5) a) pretty easy
b) i fail at lewis dot structures so failed this question
c) i talked about H-bonds and LDFs, im most likely wrong
d) dont remember
e) i said decomp because the oxidation #s didn’t change and there was no acid/base
f) dont remember
g) i said false because of the bond thing</p>
<p>6) i said it was 1st order because i thought it’s order was based on # of reactants, but it’s zero
c & d i got ****ed in the arse</p>