<p>NoxLuminus, thanks for the clarification.</p>
<p>Afruff, in the notes I have, the only exception is with aluminum. But I did Google [Fe(CN)6]4- and it does exist. I think I'll ask my teacher about that tomorrow.</p>
<p>@wxmann&noxluminus</p>
<p>rate constant k= Ae^(-{Ea}/RT) Ea is the activation energy and A is a constant.</p>
<p>it should be obvious that catalyst lowers Ea so increases the rate constant k</p>
<p>No, you misunderstand. I said that FeCN4 2+ should form according to "the rule" instead of FeCN6 4-.</p>
<p>@afruff23 it should be FeCN4 2- right?</p>
<p>@ 0C is used in gas problems for "STP" - standard temp and pressure
Standard Conditions is when you see the little superscript degree - that refers to 25degC, 1 atm for gases, and 1M for solutions. (Used for dGnaught, Enuaght, dSnaught and dHnaught)</p>
<p>so according to the "rule"
[Zn(Cl)4]2- [Ag(CN)2]- both exist?</p>
<p>"@afruff23 it should be FeCN4 2- right?"</p>
<p>I'm not him but yes</p>
<p>@ "the rule" for complex ions is just a guideline so you don't have to memorize <em>every</em> possibility. Most cations that form complex ions actually take different numbers of ligands under different circumstances. When you double the charge, you almost always get one of the acceptable answers. (Every AP complex ion question has allowed multiple answers for numbers of ligands.) Al complexed with OH- is one common exception you should know. Also Fe3+ usually only takes one SCN-, but again, the answer key usually accepts multiple answers.</p>
<p>k (rate constant) definately changes with a catalyst for the reason andy-tok mentioned. Catalysts provide a different pathway with a lower Ea</p>
<p>"so according to the "rule"
[Zn(Cl)4]2- [Ag(CN)2]- both exist?"
The "rule" doesn't tell you what exists. In Question 4, they only give you reactions that work, so you don't have to decide whether the complex ion actually exists. The "rule" just tells you that if they do work, twice the charge is a reasonably safe bet.</p>
<p>The k and catalyst thingy was on the mock for chem. I got it wrong and I believe I put that it did. It would make sense for a catalyst to lower k.</p>
<p>A little late, but official sig fig grading is -</p>
<p>1) no points deducted if you are within +/- one of the correct number of sig figs
2) max of 1 point deducted per problem for sig figs</p>
<p>and yes, you should carry all the numbers in your calculator and round at the final answer (for that part)</p>
<p>About the rate mechanism question: If a substance appears both at the beginning and end of a rate mechanism (i.e. it is not used up), then it is a catalyst. If it only appears in the middle of the rate mechanism and is used up, it is an intermediate, not a product.</p>
<p>So if one the equations section it says Fe3+ and CN- form a coordination complex you could put (Fe(CN)6)3+ ?</p>
<p>@afruff,
the total charge should 3-</p>
<p>
[quote]
So if one the equations section it says Fe3+ and CN- form a coordination complex you could put (Fe(CN)6)3+ ?
[/quote]
</p>
<p>What page is this in the pr book?</p>
<p>@andy_tok
Thanks, that's what I meant.</p>
<p>@specify
page 292 in PR 2006-2007</p>
<p>DO WE GET SCIENTIFIC CALCULATORS OR GRAPHING CALCULATOR ON FREE RESPONSE? PLEASE RESPOND TO THIS QUICKLY!!</p>
<p>Calculators allowed on part A of FR, not on part B. Any type except QWERTY is allowed.</p>
<p>Where can I get the "rules" for the complex ions? Is it in the PR or Barrons book?</p>
<p>Let me tell you exactly what PR says about complex ions:</p>
<p>"Look for trans. metal ions in solution with Ammonia, Hydroxide, Cynaide, or Thiocyanate</p>
<p>Trans metal ions form complex ions with the species above. It doesn't matter how many ligands you place on the transition metal, as long as you get the charge on the complex ion correct. In general, the number of ligands will be twice the charge on the metal ion."</p>