<p>Barricade: both arguments work as long as you supported your reason with an explanation. I put down polar bonding too.</p>
<p>hydrogen bonding doesnt work...because the N...is not next to the H..someone told me..is this true?</p>
<p>O with H is hydrogen bonding too.</p>
<p>Its the O in the water. I originally put that down but erased it and did polarity, but I can see how they both work now</p>
<p>Yatta!!
I put the same thing as you. I was pretty certain also, and with that answer I got that oxygen is the reactant in excess, because only .75 mol of the 3 O2 can react with the 3 mols of HCl since its in a 4 to 1 ratio.
I thought I was definitely correct... BUT of course it would be too nice to have gotten all 3 of equations and part Bs correct. I asked my friend what he put and then I realized that I totally forgot the rule: metal chloride + oxygen gas >>> metal chlorate.</p>
<p>Usually we think of HCl as hydrochloric acid and in that case it wouldn't work but now I realize why they said hydrogen chloride to help us remember it was a metal chloride. So I am pretty sure that the answer was 2HCl + 3O2 >> 2HClO3 and in that case HCl would be in excess.</p>
<p>I wonder if our answer will give us credit though, at least I think my part B might because it correlated with my part A.</p>
<p>If your still not sure that that is right think about this
MgCl2 + O2 >>> Mg(ClO3)2</p>
<p>Hydrogen isn't a metal I don't know if that makes a difference.</p>
<p>6a. Hydrogen bonding. Keep in mind that while the H in H-bonding has to be attached to N, O, or F, the N/O/F in H-bonding doesn't have to be attached to a H. The unbonded electron pair on the N is just begging for the H on water molecules.</p>
<p>b) Ethanol can do H-bonding and London dispersion, but dimethyl ether can only do London dispersion. Ethanol is also polar, whereas dimethyl ether isn't.</p>
<p>I think you had to have said hydrogen bonding. The AP is VERY picky about these things. You should get partial credit for dipole - dipole though.</p>
<p>6a.. Pyridine is polar, so like water dissolves like. I think I also said it had hydrogen bonding which it might but whatever. Benzene is nonpolar and polar water can barely dissolve it.</p>
<p>6b.. I said both have hydrogen bonding because for some reason I didn't think the H and the O had to be directly bonded to each other. Yeah I am stupid... so I just said that the O in dimethyl ether is shielded molecular structurally speaking and the unshielded O in ethanol allows for easier hydrogen bonding. I am not sure if that will get full credit or not because even though I don't think dimethyl ether has hydrogen bonding, it IS polar. But the correct answer was that dimethyl ether has dipole-dipole intermolecular forces which aren't as polar as the hydrogen bonding in ethanol, making its boiling point lower. Period.</p>
<p>some questions
1a) i didnt put PCo2 etc... i put [CO2] etc.. will they take off points? I excluded the solid etc so my equation was right. But i just missed the Ps. i even wrote Kp.
2b) my delta G was in joules, not in Kj, will they take off points?
4b) i got the correct answer that excess was Oxygen. But the problem is is that in my justification, i didnt use 3 moles, i used 1 mole. So i said when 1 mole of O2 is used, .25 moles of HCl are used. instead of the real answer, when 3 moles of O2 are used, .75 moles of HCl are used. Will i lose points?
6a) i said like dissolves like. So polar h20 will dissolve polar pyridine and not nonpolar benzene. Would that be accepted?
6b) i said Ethanol is polar while dimethyl ether is nonpolar. Is that sufficient?</p>
<p>Please help thanks</p>
<p>If you only put the CO and CO2 in brackets withough doing P subscript CO I dont think they will count it as right. You have to show that you are using pressure, not concentration and that doesn't prove you are using pressure.</p>
<p>For the thing about F and O....you can say that F has more protons so it has a stronger pull on its electrons.
for B i just used PV=nRT...but i know that isnt right...
How do you do 1B and 1C??</p>
<p>1b, u just put it in pv=nRt where p is 5(initial pressure b4 eq).
1c, u do 8.37(total pressure at eq) - 1.63( new pressure) = 6.74</p>
<p>can someone help me on my problems??</p>
<p>sushant,
Yeah just the brackets without P means molarity. If you put brackets with the P (which I think I did) they will accept that.</p>
<p>You can put delta G in joules or kj. Either way it was supposed to be rounded to 3 sig figs. I put 360000J which equals 360kJ. </p>
<p>Your 6a is perfectly fine. Your 6b though, is wrong I think. Dimethyl ether is polar so its forces are dipole-diplole forces.</p>
<p>Does anyone know the correct answer for 1e and how to justify it? I was in the middle of comparing the moles of carbon to something else but time ran out, so I never put an answer of increasing or decreasing. I think its decreasing though. I think I was doing the right thing though because why else would they tell you how much carbon was in there?</p>
<p>Wait why does dimethyl ether not have hydrogen bonds? Because the electrons are being pulled towards the C, not the H? </p>
<p>And can anyone please explain 6c? That seemed totally counterintuitive to me.</p>
<p>I think that was a trick. You shouldn't need it, right? Just compare the reaction quotient to the equilibrium constant.</p>
<p>its decreasing. All you do is Q = (2^2)/2 = 2
since 2 < Kp (27.something), the reaction goes to the right decreasing CO2
can someone help me out with my problems? They are on page 5!</p>
<p>I think problems 2 and 3 were ridiculously easy. I got full credit on problem 2 and for problem 3 I might miss one point on part c only. How were you able to find out what delta S could have been if delta H wasn't given? I assumed delta H was negative because it gave off enegry according to the electrical potential.</p>
<p>Oh, I didn't even think of Q. Thats mad easy.</p>
<p>I answered all of your questions sushant. You just didn't say what reaction you put for 4b so I don't know. But as long as you said what the correct excess reactant was it shouldn't matter.</p>