Recent content by aleksbooker

Mmk. It took me a minute to get back to this, but here's my understanding. Strong acids disassociate completely not necessarily because their components are held "more tightly" in compounds with the solvent particles, but rather because the compounds they form with the solvent particles (or the...

This is awesome. I'm learning so much (or at least, I think I am). Here's what I've gathered from you guys so far: HCl and other compounds that dissociate completely in solution are called strong acids because of this behavior, not the other way around. This behavior (complete dissociation) is...

Hmm... I'll save that for a later date. I want to know, but it's not essential to this class and I unfortunately can't spare the time to research. Hopefully I'll run into it again later, though. Appreciate your help guys!

So, by nature, the molecules of strong electrolytes have *weaker* holds on their protons? That way more of them can disassociate into solution and actually act as electrolytes. And more, strong acids have weaker holds on their protons (H+ ions), while weak acids have stronger holds on their...

I'm confused about the dissolution of strong acids (like HCl) in water. We're told that they dissolve completely because They are strong electrolytes. The hydrogen atom that is lost during dissociation is *not* strongly bound to the rest of the acid molecule. Therefore, the solvent...

Yea, L was the length of the string.

Mind blown. Thanks @Doc Al. Okay, that works - I'm assuming it only makes to do something like this when you can use it to eliminate a variable (like T in this problem).

How did they "divide the equations"? Homework Statement Two 5.0g point charges on 1.0m-long threads repel each other after being charged to +100nC. What is angle theta? You can assume theta is a small angle. Homework Equations K=9.0*10^9 Nm^2/C^2 g=9.8m/s^2...

@pasmith, thanks for your response. I forgot how much I hate working with e. Thanks for pointing out how simple that really was. Here's what I was doing: S = \frac{1}{2}\pi \int (e^{2y} + 2 + e^{-2y}) dy Then, breaking it up into three separate integrals and working with just the...

Homework Statement Find the area of the surface generated by revolving the curve x=\frac{e^y + e^{-y} }{2} from 0 \leq y \leq ln(2) about the y-axis. The Attempt at a Solution I tried the normal route first... g(y) = x = \frac{1}{2} (e^y + e^{-y}) g'(y) = dx/dy = \frac{1}{2}...

Oh. That's a great explanation. Why didn't they put *that* in the textbooks?

Edit: I've made some progress on this one, and I now understand that the valence and conductance bands are composed of molecular orbitals contributed by each metallic atom joining the molecule. For example, three lithium atoms would contribute three total molecular orbitals, resulting in three...

Cool. Thanks for confirming/clarifying. :)

Should I then mentally extend the top mirror beyond the corner so that it "sees" the reflection from the bottom right mirror?

Homework Statement Two 3.0m wide mirrors meet at a corner. Taking the corner as the origin of the x/y axis, A red ball is placed at point A (-1m, -2m). 1) How many images are seen by an observer at point O? [Point O is not given coordinates, but looks to be at approximately (-3m, -3m)]...