<p>How’d you guys explain that Trial #1 was inconsistent?</p>
<p>@collegedreams29</p>
<p>I don’t remember what I put for G=-nFE</p>
<p>But for 5a…I put that since Temperature is directly proportional to the average kinetic energy of the system and KE=(1/2)mv^2 and the mass of the particles are not changing, the velocity HAS to increase, thus the particles are moving faster and have a greater probability of colliding with themselves and with the sides of the container. I also put that since the volume wasn’t changing, the pressure had to increase.</p>
<p>@collegedreams29</p>
<p>I’m not sure about this, but I found the change in temperature of each trial and compared it to the mass of MgO in the system and because the mass of MgO was doubled between 0.25 and 0.5 grams and the coefficient of it was 1, then the change in temperature had to double between 0.25 and 0.5 grams. Trial 1 had only 1 degree of Temp. change while trial 3 with the same amount of MgO had 2 degrees and Trials 2 and 4 (0.5 grams) had a 4 degree change</p>
<p>What did you guys get for the net-ionics (#4)???</p>
<p>Yes the limiting reactant IS MgO. My mistake was in part c, and the solution I gave should be the correct one. And its 96,500 collegedreams.</p>
<p>For the calirometer one, I did not add to .25g to the mass of the water, however because of sig figs, the answer is the same. A point should still be deducted right?</p>
<p>What did you guys put for #2 (d)??</p>
<p>@randomandweird,</p>
<p>electrons need to be in the liquid state to move around and conduct electricity? solid, they would essentially not be able to conduct at all.</p>
<p>3Ca+2 + 2PO4-3 = Ca3(PO4)2</p>
<p>CO2 + H2O = H2CO3</p>
<p>Zn + 2H+ = H2 + Zn+2</p>
<p>@ randomandweird
lol I got that one wrong. I said that something about covalent structures and limited conductivity as a solid or something.</p>
<p>2c was it PV=nRT?</p>
<p>@collegedreams29
Yes that’s what I used.</p>
<p>Did you guys get 26.1 for 2.a, and nf=AS for 2.b?</p>
<p>5B anyone? I totally did this wrong. rate law for 5C is just first order to both reactants?</p>
<p>Thanks collegedreams29 and darkaeroga</p>
<p>I think that’s what I put…but my memory’s going down on me…so not too sure…</p>
<p>@FlyingWombat</p>
<p>For 2(b)</p>
<p>I used C=It
Not sure if that’s right</p>
<p>For 5B I said that as the reaction proceeds, equilibrium shifts to the products decreasing the number of moles of gas. As the number of moles decreases, the force and frequency of collisions with the container decrease.</p>
<p>What’s 6(b)?</p>
<p>I think it’s an ether with 2 methyl groups attached to the O.</p>
<p>For 6B, I said the answer was Compound 2, since the compound in Box X exhibits Hydrogen bonding, and would have stronger intermolecular forces, which would cause it to have a higher boiling point compared to the compound in Box Y which was nonpolar due to its symmetry.</p>
![http://www.chemistry-reference.com/images/structural/dimethyl%20ether.png[/url]](http://www.chemistry-reference.com/images/structural/dimethyl%20ether.png%5B/url%5D){kind=link}
