<p>Answer check on 1(d) and 1(e)?
For 3(b) how do you know how many moles to multiply by in the equation?
For 3(c) I'm pretty sure dS increases due to the fact that it produces a gas and liquid, and thus is more disorderly.
Answer check on the reactions/answers to questions?
For 5(b-c) I discussed how the more electrons each atom has, the more electron repulsions and thus the more energy it requires to remove one electron.
Answer check on 5(e:i,ii).
5(f) Polar due to the lone pair.
6(a) Polar.
6(b) Hydrogen Bonding.
Answer checks on 6(c-d)</p>
<p>Wasn't sure on lab question if someone would be willing to explain.</p>
<p>I think it should be Al(OH)3 + OH- -> Al(OH)4-. The trick here is recognizing that Al(OH)4- still has 6 ligands because it i really Al(H2O)2(OH)4-, thus satisfying the # of ligands = 2*charge rule. I checked this in my chemistry textbook.</p>
<p>Right, murderah, but HCl is a polar molecule. Cl2 is nonpolar, and nonpolar molecules tend to have weaker intermolecular forces -- London dispersion v. dipole-dipole. So I don't get why Cl2 would boil at a higher temperature.</p>
<p>Cl2 has greater polarizability due to its larger size and greater number of electrons, meaning its induced/instantaneous dipoles will be stronger, so its LDF are pretty strong, probably stronger than the hydrogen bonds between HCl molecules?</p>
<p>So London dispersion forces can really be that strong? My memory was that because they were instantaneously induced, they wouldn't be as strong as dipole-dipole forces which are permanent (and the electroneg is .9, which seems pretty substantial).</p>
<p>im pretty sure i got all the net ionic equations right. I even confirmed with my teacher. 4a) Al3+ + 6OH- = Al(OH)6^3-
4b) 4HCl + O2 = 2H20 + 2Cl2
4c) nobody seems to have difficulty with that.
for 4b), my teacher says chlorites/chlorates dont work because we are dealing with HCl gas.</p>
<p>I think Al(OH)6^-3 will also be accepted. Thats what I put.</p>
<p>And yes I always thought of LDF automatically being the weakest of the intermolecular forces. But I just agreed with what they told me, that Cl2 had a higher boiling point than HCl. So I wasn't going to argue against it. The only thing I could think of for that being the case if because of the numerous electrons and its induced dipoles will most likely be stronger than HCl which probably has weak dipole bonds.</p>
<p>That would be amazing if chlorates don't work because of HCl gas. Actually, your teacher is probably right. Chlorates are usually from a solid and oxygen gas. </p>
<p>In that case I got every single equation right. Haha! Thats what I thought when I was taking the test anyway. =)</p>
<p>And just so everyone knows number 4c is this
K2O + H2O >> 2K+ + 2OH-</p>
<p>Oh yeah, for 4b I got what you got sushant, but by mistake I broke up HCl into its ions (doh! no water).. so I had 4H+ + 4Cl- + O2 as my reactants.. Do you think I will get full credit still?</p>
<p>and can anyone explain ... basically all of #6 :p</p>
<p>i messed up the shape of XeO3 based on my wrong diagram...would that be partial credit</p>
<p>and if i just said "SiO2 interacts and bonds with other SiO2 molecules so it is harder to break those bonds and so the BP is higher than SO2, where the molecules don't interact with each other" would the be partial credit?</p>