<p>ya i think i finally get the buffer questions now. the answer is always the weak electrolyte and the conjugate(thanks jamesford for pointing that out) So in the problem i asked before everything is else is strong and will dissociate so that leaves B</p>
<p>^ Judging by some of my practice tests i think there is a high chance of a titration FRQ. The one I did today was strong base/weak acid one</p>
<p>hmm do you know the amp?</p>
<p>velleity, the trick is you have to find the amp. I’m working it now</p>
<p>although you can do the whole electroplating out the ratio of mole electrons is 2 is to 3
multiplying 243 by 2/3 is approx. 180.</p>
<p>ooo that’s so much easier than what I was doing. I need to learn how to do quick MC math. I need to realize that they can’t give you a 5 minute problem for 1 question (well I hope :P)</p>
<p>HgO(s) + H2O <–> HgI42– + 2 OH–
Consider the equilibrium above. Which of the following changes will increase the concentration of HgI42– ?
(A) Increasing the concentration of OH–
(B) Adding 6 M HNO3
(C) Increasing the mass of HgO present
(D) Increasing the temperature
(E) Adding a catalyst</p>
<p>The answer is B, could anyone explain this? Thanks</p>
<p>wait can someone still go over this problem again? i don’t get it. the shortcut of mole electrons and the other way. A steady electric current is passed through molten MgCl2 for exactly 1.00 hour, producing 243 grams og Mg metal. If the same current is passed through molten AlCl3 for 1 hour, the mass of Al metal produced is closest to</p>
<p>a) 27.0 g
b) 54.0 g
c) 120 g
d) 180 g
e) 270 g.</p>
<p>it’s D. WHY?</p>
<p>If you pass the same current, but you need ot move 3 electrons for one <em>plate</em>, then you would plate 2/3 slower, leading to 180 g.</p>
<p>HNO3 is a strong acid and adding it will dissociate when added to the solution, resulting in H+ and NO3- ions. The H+ would react with the OH- on the right side of the reaction to form water. Since the OH- concentration is being reduced, the reaction will shift to the right and more HgI42- will be produced.</p>
<p>Oh, I get it. Thanks farlad.</p>
<p>When do we apply this formula?pH = pKa + log [A-]/[HA]
I’ve never used it before…</p>
<p>For buffer solutions, when you’re trying to find the change in pH.</p>
<p>whaat? lol sorry i still don’t get it. can you write out the equations that occur with the electrons in them/ where the 2/3 comes from? sorry i havent looked over these types of problems too much yet.</p>
<p>This might seem silly but I just cannot get what to do exactly…</p>
<p>Approximately what mass of CuSO4-H2O (250 g/mol) is required to prepare 250 mL of 0.10M copper (II) sulfate solution.</p>
<p>A) 4.0g
B) 6.2g
C)34g
D) 85g
E) 140g</p>
<p>The answer is B</p>
<p>what’s the formula i would need for that electroplating stuff?</p>
<p>is it just:</p>
<p>mole e- = (amperes x seconds) / 96500C?</p>
<p>number of e- = current x time / (charge per e-)
then divide by avogadro’s number to get moles of e-</p>
<p>Since the current and time are the same, the same number of moles of e- are produced. Since it takes 2 e- to produce 1 Mg whereas it takes 3 e- to produce 1 Al, the amount of Al will be 2/3 the amount of Mg when subjected to the same current for the same amount of time.</p>
<p>Terrence… Let’s see.
there’s .025 moles of CuSO4, and one mole of that hydrate is 250.
therefore, 0.025 x 250 = 6.2g</p>
<p>doesn’t my formula account for moles already? Does anyone else know that formula?</p>
<p>The moles will be .250 * .10 M and that is 0.025 moles. Now you know the molar mass is g/ mol so multiply those two together to get the neccessary grams, and you get the answer.</p>