A few chem questions!

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1

<p>Ok, should I know the Henderson-Hasselbach thing for buffers for chemistry? And if so, could someone please explain it?</p>

<p>What's the diff between end point and equivalence point? They seem to be the same to me...</p>

<p>do alpha particles have a charge, considering they are neutral? (a practice question in PR asked which decay particles had a charge, and alpha was one of the answers).</p>

<p>how is molality used in the calculation of boiling point elevation and freezing point depression? i thought that you just need to see how many ions the thing dissociates into? i don't know what to do with molality</p>

<p>thanks!</p>

2

<p>too lazy to answer all of them</p>

<p>but alpha particles MUST have a charge...they are not neutral.......</p>

<p>alpha particles are the equivalent of helium nuclei, not helium atoms</p>

<p>electrons are not inluded</p>

<p>thus, all u have is the nuclei of the helium with 2 protons and 2 neutrons
so the charge of the alpha particles is +2</p>

<p>i think thats actually how the nucleus was discovered or something...</p>

<p>the particles being shot through a gold sheet and bending downward toward the negative due to charge interactions?</p>

<p>well hope that answers ur second question</p>

3

<p>
[quote]
how is molality used in the calculation of boiling point elevation and freezing point depression? i thought that you just need to see how many ions the thing dissociates into? i don't know what to do with molality

[/quote]
</p>

<p>It's basically the same thing. Freezing point depression=(i)(m)(Kf). For example, let's say you put 0.2 mols of NaCl into 100g of water. The molality would be (0.2 mols/0.1 kg solvent), or in other words, 2 molal. Then, "i" is just the number of particles it dissociates into: in other words, Na+ and Cl-, so two. Your final equation would be "fp depression=(2)(2)(Kf)" where of course Kf is the constant for water.</p>

<p>If you put, instead of NaCl, CaF2, then "i" would be 3 since it dissociates into Ca(2+) and 2 F-, changing your equation to "(3)(2)(Kf)"</p>

4

<p>Thanks Goldshadow. Is using molality the same as just multiplying the number of ions dissociated x the depression thing?</p>

<p>Like for putting NaCl on ice, since NaCl goes into two ions, I just say 2 x 0.51 for boiling point elevation and 2 x 1.86 for freezing point depression. Is that the same as your case where you multiply some constant (what is the constant, anyway?) by 4?</p>