<p>bumpity mcbumpalot.</p>
<p>what’s the tyndall effect?</p>
<p>
</p>
<p>This is what i have, it’s probably not right but hey at least i tried right?</p>
<p>part A:
R: CH3NH2 + H20 -> CH3NH3+ + OH-
I(initial): .225 - 0 0
C(change): -x - +x +x <-stoich ratio
E(end): .225-x - x x</p>
<p>Kb=[(CH3NH3+)(OH-)]/[CH3NH2]=5.25e-4
Kb=[(x)(x)]/[.225-x]=5.25e-4
x=.0106M</p>
<p>note:volume does not matter(pH-wise) so we can simply use a 1L sample for simplicity.</p>
<p>Part b:
same RICE but instead the Initial of CH3NH3+ is .01/.120(to get concentration) instead of 0. So plugging in for the Kb expression and solving for x(note: you MUST use the quadratic formula when solving for x since the Kb and initial are so close), you get x=1.42e-4. -log[OH-]=pOH -log[1.42e-4]=3.8478 pOH + pH = 14. pH=10.15</p>
<p>Part C: in order to get a pH of 11, we must have a pOH of 3. 10^-3=.001 gives us the number of OH- ions. Leave that where it is. We also know that our Kb expression is: [(.01/.12+x)(b/.12+x)]/[.225-x]=5.25e-4 where b is the number of initial moles that we need. This is where the number of OH- ions come in.
[(.01/.12+.001)(b/.12+.001)]/[.225-.001]=5.25e-4. Solving for b, we get b = 4.73 moles</p>
<p>part d:
when you add more water, you stress the left side so according to Le’Chatlier’s principle, the reaction will proceed to the right. This means that there are less H+ ions present and more OH- ions are present. This means that the concentration of OH- will go up and H+ will go down so the pH will go up.</p>
<p>oh ya the tyndall effect, wasn’t that the scattering of light in a solution? hmmm i remember something about being able to tell the difference between a suspension and a true solution with it.</p>
<p>(really easy)
What does Hess’ Law state?</p>
<p>Hess’s law: it doesn’t matter if enthalpy for a rxn is calculated in one step or a series of steps because enthalpy changes are state fxns.</p>
<p>Describe the molecular geometry of CF4 and SF4, and how the polarities of the two differ (in terms of their geometry). [real ap FR question]</p>
<p>CF4: tetrhedr0n symmetrical so polarity is canceled out
SF4: see-saw polar because there is one pair of electron in the left part. It distorts the shape and makes it polar.</p>
<p>CF4: Tetrahedral, therefore: No net dipole.</p>
<p>And in response to VDW/LDF/dipole forces, you can just use the terms LDF and dipole-dipole and not worry about VDW.</p>
<p>GOOD IDEA!! haha. :)</p>
<p>What is the substance called that is produced in a step of a rxn but then consumed in the next step, so is neither a product or reactant?</p>
<p>intermediate</p>
<p>which has the larger density?
Na or Ne?</p>
<p>Which has a larger atomic radius? Li or F?</p>
<p>Li if they’re on the same period.</p>
<p>Ne, cuz its more tightly packed?</p>
<p>Ok, here is some Barron’s stuff…don’t know if it is needed.</p>
<p>Why can a buffer solution be made from a strong acid and it’s salt? I don’t understand that.</p>
<p>How does a spectrophotomer work? What law does it correlate with, and what does this law say?</p>
<p>A spectrophotometer measures absorbance or transmittance of a liquid. It relates to the Beer-Lambert Law.</p>
<p>I remember doing a prac test the other day that had a question about that. So, um, what are the applications of a spectrophotometer?</p>
<p>
No, scattering of light in a colloid. You can tell differentiate between a colloid and a solution with it.
What is Brownian motion?</p>
<p>ok stop asking irelevant questions.
Let’s do some reaction problems.
first easy ones
SO2 gas is added into water
HCl gas is added into a KF solution.</p>