<p>On some practice tests I did for similar problems, you had to mention Q in order to get full credit.</p>
<p>You can tell the sign on delta S just by looking at the chemical equation.</p>
<p>the n in pyridine isn't bonded to hydrogen lol but the hydrogen in water is........... meaning the nitrogen will hydrogen bond. so hydrogen bonding is correct.</p>
<p>Yeah I know you could but usually its off of more simple equations like going from a solid and a liquid to gas or something. My teacher said never worry about finding delta S solely off the equation for something as complicated as this one where it goes from solid and aqueous to gas, liquid, and aqueous. If you had to do that, did you count the H+? So it would be going from total of 13 moles to 9 moles and therefore the sign of the delta S would be negative as it gets more orderly.</p>
<p>predicting the sign of s will only be one point. who cares if you got it wrong or not haha.</p>
<p>Haha, then I got my 5.</p>
<p>think about it like "number of possible positions for the molecule to go to"</p>
<p>I looked at aqueous as liquid, it is definitely less disordered than gas. hope that was okay</p>
<p>ah i put more disordered! meaning + delta H. I say aq -> gas and solid -> aq so i pretty much summed up that it had to be positive and disorderly!</p>
<p>do any of you know 5e?</p>
<p>sp3d2 and trigonal pyramidal</p>
<p>For the third reaction, I put: </p>
<p>HCl + O2 --> H2 + ClO2....What did you guys put?!?!</p>
<p>I put that same thing. Im pretty sure thats right....</p>
<p>I thought that the rate law problem and the one where you had to draw the lewis structures were both gimmes....like the entire problems.
Like I have seen preatty much those exact same questions before in practices.</p>
<p>^^ there hcl would be reduced because the h goes from +1 to 0. I put 2HCl + O2 --> CL2 + 2H20 because the CL goes from a -1 to 0, therefore oxidized.</p>
<p>
[quote]
^^ there hcl would be reduced because the h goes from +1 to 0. I put 2HCl + O2 --> CL2 + 2H20 because the CL goes from a -1 to 0, therefore oxidized.
[/quote]
That's right, I made the mistake of confusing oxidizing and reduction.</p>
<p>I seriously hope 4B is 4HCl + O2 -> 2H2O + 2Cl2, cuz I had completely forgot about chlorates, chlorites, and all that crap. Besides, oxidation and reduction do occur anyway, right?</p>
<p>What did you guys say for 6d, where it says the boiling point of Cl2 is higher than that of HCl. Kind of against my beliefs, I blamed it on London dispersion forces, how Cl2 has more electrons than Hcl. Not sure it's right. Kind of worried about 6a now, with all of this hydrogen bonding thing... I just saw that lone pair as, "Oh! It's polar!"</p>
<p>It's sad, but I was kind of dissapointed that I didn't get to use ICE.</p>
<p>Thats what I put to thatonestudent. I used ICE though in problem 1.</p>
<p>isnt trigonal pyramidal sp3?</p>
<p>You're probably thinking about trigonal planar. Trigonal pyramidal is a tetrahedral with a lone pair of electrons.</p>
<p>for Cl2 and HCl thing, i said that generally, the polar molecule like HCl would have a higher boiling point than Cl2, but when the mass difference between molecules is great, the LD forces can overcome Dipole-Dipole. Since Cl2 is double the size of HCl, the LD forces in there overcame the dipole-dipole in HCl.</p>
<p>but isnt the hybridization for tetrahedral and trigonal bypryamidal sp3?</p>