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<p>You’re right. The negative charge on HSO4- makes it very difficult for the positively charged proton to dissociate.</p>
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<p>It’d be a good idea. As for the what element’s in what, know what metals are present in certain alloys. Not sure what else.</p>
<p>Do we need to know how to calculate voltages of voltaic cells?</p>
<p>^I don’t think so. It’s pretty simple, though. Just write the voltaic cell reaction and sum the potentials. Coefficients don’t matter. If they do have such a question, they will give you the potentials in the problem, as you aren’t provided with a table of standard potentials.</p>
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That seems logical. It’s what I’ve been doing, anyway.</p>
<p>How do u guys compare ap chem and sat chem??? is sat chem lot easier??</p>
<p>Now that we only have 3 days left, how should I spend this time reviewing? I’m basically done with Barron’s and I’m going to review the CB practice test I just took.</p>
<p>would someone be so kind as to explain hybridization to me? I’ve read stuff about it in multiple books, but I just can’t seem to understand it fully.</p>
<p>All you really have to know is that s, p, and d orbitals “hybridize” (combine) to form new electron orbital structure. To find the hybridization of a certain atom, count the number of electron areas around it (single bonds, double bonds, triple bonds, and unbonded pairs all count as one electron area).</p>
<p>1 area = sp hybridization
2 = sp^2
3 = sp^3
4 = dsp^3
5 = d^2sp^3</p>
<p>Would someone care to explain #62 from the official released test?</p>
<p>Jamesford, could you list the major alloys? or perhaps do you know where I can find a list of alloys that will be on the test?</p>
<p>and where’s the official released test, so I can look at #62</p>
<p>And for hybridization remember that H2O and NH3 are sp3 hybrids because they have the electron pair geometry of a tetrahedron, meaning their unpaired electrons take the places of where the atoms would be if it was a tetrahedron. Here’s a good review of all of this stuff:
[VSEPR</a> - Molecular Models](<a href=“http://chemlabs.uoregon.edu/GeneralResources/models/vsepr.html]VSEPR”>http://chemlabs.uoregon.edu/GeneralResources/models/vsepr.html)</p>
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<p>Brass: mixture of copper and zinc
Sterling silver: mixture of silver and copper
Steel: mixture of iron and carbon
Bronze: mixture of copper, zinc, and other metals
Pewter: mixture of tin, copper, bismuth, and antimony</p>
<h1>62 asks what intermolecular forces make it possible to liquefy hydrogen gas. The choices are London dispersion, covalent bonding, ionic bonding, dipole-dipole attraction, and hydrogen bonding.</h1>
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<p>Think of this in terms of equilibrium and LeChatelier’s principle. H2SO4 fully dissociates, giving 1.2 moles of H+ ions. When these react with the hydroxide ions of NaOH, a stress will be placed on the system and the remaining HSO4 ions will continue to dissociate in order to attain equilibrium concentrations.</p>
<p>Jamesford, that question is particularly difficult. At this point, you should know that hydrogen gas is H2, a purely covalent, and thus unpolar, molecule. You should also know that intermolecular attractive forces determine (to a certain extent) the melting and boiling points of any given substance. For instance, polar molecules will have greater attractive forces amongst themselves, causing relatively high boiling points (the energy needed to overcome the attractive forces and release molecules into a gaseous state is greater).</p>
<p>Now, you could go through this question and eliminate the incorrect answers. Let’s start with answer B, because it is incorrect. Dipole-dipole attraction is caused by polarity and thus exists in polar molecules. (The positive dipole of one molecule will attract the negative dipole of another.) Hydrogen gas isn’t polar, and thus has no dipole-dipole attraction.</p>
<p>Answer C: covalent bonding is also incorrect, because, although the hydrogen atoms in hydrogen gas are attached by a covalent bond, this bond only affects the two atoms in question (it is an intramolecular force), and thus doesn’t affect a given molecules relation to another.</p>
<p>Answer D: H2 doesn’t have an ionic bond, and this answer is automatically incorrect.</p>
<p>Answer E: this is probably the trick answer. Hydrogen bonding only occurs when a hydrogen atom is attached to an atom of either Fluorine, Oxygen, or Nitrogen. This creates a polar bond, such that the hydrogen (the positive dipole) will form an intermolecular bond with the F, O, or N atom of another molecule (the negative dipole). This is responsible for the high boiling point of water; the hydrogen atoms (positive dipoles) are attracted to the oxygen atoms (negative dipoles) of other molecules. Since H2 isn’t polar, and since the hydrogen atoms aren’t attached to F, O, or N atoms, hydrogen bonding is ruled out.</p>
<p>This leaves us with choice A, London dispersion forces. These generally exist in every substance. The electrons in one molecule (or atom, when dealing with inert gases) will repel the electrons in another molecule, creating what is called a “dipole moment” in the second molecule. This dipole is instantaneous, but is still responsible for a certain amount of intermolecular force. It is byfar the weakest type of intermolecular attraction, but is the only one present in hydrogen gas.</p>
<p>Note that if a given substance had absolutely no attractive forces holding its molecules/atoms together, this substance would most likely be an ideal gas, and would be impossible to liquefy, let alone solidify.</p>
<p>This is a particularly tricky part of chemistry, so I don’t know how helpful my answer was.</p>
<p>^right, but then it would be NaOH + HSO4-, which will result in a basic solution… Perhaps the question means “neutralize” in terms of completely getting rid of H2SO4, and with no regards to the pH?</p>
<p>@Jamesford… thanks for the list. Also, of the choices, H2 gas only has covalent bonding and london dispersion. If you want to liquify it, you need strong IMFs, and so I would go with covalent bonding.</p>
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<p>No, the addition of NaOH (and thus of hydroxide ions) will shift the position of the equilibrium, and HSO4 ions will dissociate until equilibrium concentrations are attained. It’s a fact.</p>
<p>Whats the brakedown on the percentages of questions?</p>
<p>Specifically, descriptive chem, which is not taught in school.</p>
<p>Also, are there any hard acid/base problems or just straightforward with logs?</p>
<p>yea jamesford i struggled with that question too but the way rockermcr put it was awesome. i think i recall in one of the practice tests something about how gold resists reacting with an acid…do we need to know these things?</p>
<p>@ rocker: thanks for the detailed response. I did have it narrowed down between covalent and London dispersion, but I just guessed wrong.</p>
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<p>Yes. Gold and platinum tend to be very unreactive and exist in nature by themselves (i.e. not as part of a compound). The alkali metals are very reactive and exist almost exclusively in compounds. The other metals fall in between the two extremes.</p>
<p>Platinum is actually used as a catalyst in many organic reactions.</p>
<p>rocker, I know that the addition of NaOH will cause more H+ to dissociate from HSO4-, but the solution will be basic, since the conjugate base (SO4-2) is strong. </p>
<p>Perhaps I’m misunderstanding your explanation?
I still think H2SO4 will completely dissociate into HSO4- + H+, so I will need at least 1.2 moles of NaOH, but HSO4- is a weak acid. So would “neutralizing” mean getting rid of all H+, including the ones attached to HSO4-, or neutralize as in get as close as possible to pH of 7?</p>