<p>Well, not too late to get an indication of how ready you are on the eve of the test. I'll take a released if anybody has one. If not, w/e. I'm pretty confident that I can five this....</p>
<p>i will get a 5 I WILL GET A 5 I WILL GET A 5 ASDFJKL;KJDFKLASJF;SJFLKASJFSJ;FLJASDFJ;DJFLDJ
not</p>
<p>anyone want to teach me wat pKa and pKb are and what they do etc?
please??????????????????</p>
<p>pKa/pKb is just -log(Ka/Kb)</p>
<p>i hate no.4.....................</p>
<p>Write a balanced net ionic equation for the spontaneous reaction that takes place in the cell.
Zn2+(aq) + 2 e− → Zn(s)
Ag+(aq) + e− → Ag(s)
Zn(s) + 2 Ag+(aq) → Zn2+(aq) + 2 Ag(s)</p>
<p>ok... how on earth do you know this is the way the reaction goes? I'm sure it has something to do with the E knot values, but idk</p>
<p>the only use of pKa the acid dissociation constant and pKb is the base dissociation constant</p>
<p>michael, you see which one is most positive and that's the cathode. do cathode - anode and if it's negative then it's spontaneous. If not, then you switch the values (meaning you switch the cathode and anode with one another). Tell me if that's not eloquently written enough.</p>
<p>Hey narcissa since i know youre gona check this thread can you respond to my PM at some point?</p>
<p>pKa is mainly for use in the Henderson-Hasselbach equation which is </p>
<p>pH=pKa+log([base]/[acid]). The base and acid part are just [A-] and [HA]. This equation is mainly for buffering solutions. I just learned this myself....</p>
<p>Find the mystery resultant and its concentration after equilibrium</p>
<p>C2H4OH + OH = H2O + C2O + ____
1.24x10-2</p>
<p>K = 1.3 x 10-3</p>
<p>Scratch the previous post... too easy..</p>
<p>LOL TRY THESE SUCKERS (BTW I GOT A 5 LAST YEAR)</p>
<ol>
<li><p>Malonic acid (HO2CCH2CO2H) is a diprotic acid. In a titration of malonic acid with NaoH, stoichiometric points occur at pH = 3.9 and 8.8. At 25.00-mL sample of malonic acid of unknown concentration is titrated with 0.0984 M NaOH, requiring 31.50mL of the NaOH solution to reach the phenolphthalein end point. Calculate the concentration of malonic acid in the unknown solution.</p></li>
<li><p>Consider a concentration cell that has both electrodes made of some metal M. Solution ! in one compartment of the cell contains 1.0 M M2+. Solution B in the other cell compartment has a volume of 1.00 L. At the beginning of the experiment, 0.0100 mol of M(NO3)2 and 0.0100 mole of Na2SO4 are dissolved in solution B (taking account of volume changes), where the reaction </p></li>
</ol>
<p>M2+ (aq) + SO4 2- (aq) = MSO4 (s)</p>
<p>occurs. After precipitation of MSO4 is complete, the cell potential is found to be 0.44 V at 25 C. Assuming that no other redox process occurs in the cell other than the concentration cell process, calculate the value of Ksp for MSO4 at 25 C.</p>
<p>GOOD LUCK!!! </p>
<p>PM ME IF YOU WANT MORE PRACTICE </p>
<p>P.S. IF YOU CAN SOLVE THESE RELETIVELY EASILY I SAY YOU WILL GET AT LEAST A 5 (YES, A 5)</p>
<p>None Of That Stuff Was Actually On The Test</p>
<p>Which of the following atoms contains exactly two unpaired electrons?
S, Ca, Ga, Sb, Br</p>
<p>I got this problem today in a practice test and was wondering if there's a fast way of doing it without sitting there and drawing the entire thing out...</p>
<p>it would be sulfur because P orbital has 6 electrons and S has 4 electrons in P orbital easily seen on the periodic table so 2 of them would be paired together and 2 of them would be unpaired</p>
<p>drawing it out? just look at the periodic table and look at what is in what orbital.. np^4 orbitals are usually the cases where there are 2 unpaired electrons</p>
<p>harder to tell with d orbitals because even though nd^8 would imply 2 unpaired, the atom would probably snag the 2 electrons from its (n+1)s^2 orbital and put them into the d-orbitals to make it nd^10</p>
<p>sulfur has 2 unpauired e-... 3p^4, so put one e- in each orbital, then pair.. so only one extra gets put in to make 1 pair, so 2 unpaired e- left</p>