Descriptive Chemistry + Basic Inorganic Chemistry Educational Game

<p>I guess the heat released by the exothermic process increases the entropy of the surroundings, which decreases the entropy of the system. That is why this reaction only occurs at LOW temp.</p>

<ol>
<li><p>why does the complex ion cause a color change?</p></li>
<li><p>What determines how viscous a fluid is?</p></li>
</ol>

<p>Now for me to ask you all a question that i have lol. How did you know hydrogen peroxide was powerful?</p>

<p>Usually, polyatomic ions or molecules with alot of oxygen are good oxidizing agents.</p>

<p>ok in that perspective yeah.
Some complex ions have colors like [Fe(H20)6]3+, [Cu(H2O)6]2+. the color is because of the d orbital.
viscosity is determined by the intermolecular force. If it is stronger it defies gravity better. H2O2 changes into H20 + O2 basically losing some Oxygen which means it is reduced, which means it's a oxydizing agent. The process is very abrupt.</p>

<ol>
<li> Intermolecular forces. I heard that glass is an extremely viscous fluid.</li>
</ol>

<p>You guys are probably way ahead of me, but on a quantitative basis, take Gibbs free energy equation: G=H-TS
G must be negative for a spontaneous reaction to occur. So if H is neg (exo) and entropy is also negative, then the only way for G to be negative is if T is very low.</p>

<p>What kind of bondings are present these two molecules?</p>

<p>Solid ethyl alcohol, C2H5OH
and
Silicon dioxide, SiO2</p>

<p>choices:</p>

<p>a network solid with covalent bonding
a molecular solid with zero dipole moment
a molecular solid with hydrogen bonding
an ionic solid
a metallic solid</p>

<p>[real ap question btw]</p>

<p>Name a crystal with that is highly electrically conductive and why.</p>

<p>Solid ethyl alcohol, C2H5OH - hydrogen bonding
and
Silicon dioxide, SiO2 - network covalent bonding</p>

<p>i'd go with network covalent?</p>

<p>Trivia [real ap Q]:</p>

<p>Which is used to etch glass chemically?</p>

<p>For the production of fertilizers?</p>

<p>choices:</p>

<p>Hydrofluoric acid
Carbon dioxide
Aluminum hydroxide
ammonia
hydrogen peroxide</p>

<p>o..seperately. i was gonna say... so</p>

<p>so first is hydrogen bonding, and by POE, network covalent</p>

<p>Wow, that's a rediculous question. I am not feeling good about this exam. I'd guess Aluminum hydroxide for first and H2O2 for the second</p>

<p>ANyone know the crystal one??</p>

<p>Hydrofluoric Acid for etching glass</p>

<p>Ammonia for production of fertilizers <--- commonly asked question</p>

<p>For the crystal one, I'm guessing graphite if you consider it as a crystal. It has a sea of electrons floating around being shared with all the carbons, which enables the free floating electrons to conduct electricity. If graphite is not a crystal, then i dunno. lol</p>

<p>Hey, can you guys help me out on this one. How would you rank these forces in order from weak to strongest: Van der waals, ionic, covalent?</p>

<p>^ is that right??</p>

<p>never heard of Van der waals... is that the london forces?</p>

<p>Van der waals include dipole dipole and london dispersion forces. Relatively weak.</p>

<p>ionic covalent van der waals(synonym for london force)</p>

<ol>
<li><p>Van der waals forces include the intermolecualar forces: london, dipole-dipole.
And i meant to include the intermolecular force of Hydrogen bonding.
I know strengths increase from London-dipole-hydrogen, but what about the other kinds?</p></li>
<li><p>For the crystal one, i was looking for metallic crystals. for some reason, when these atoms bond, their valence orbitals get delocalized, and electrons can flow pretty freely.
However, Graphite, a network covalent, is also a good conductor. Nice job! See you probably knew that cuz you're smart. I just know it from studying this past week. What's interesting is that diamond is also a network covalent, but is a poor electrical conductor.</p></li>
</ol>

<p>Oh yea! I think "delocalization of electrons" is key to the question you asked. It's nice to remember important phrases. More partial points on the FRQ.</p>