<p>@rawk: Ah that makes sense. But I thought nitrate were 2 double bonds and 1 single???</p>
<p>you might be thinking its a double bond but thats bc its resonating its actually single</p>
<p>There is too much bad Chemistry going on around here right now. I’m going to sleep.</p>
<p>By the way, </p>
<p>Bond order = (# of bonds formed)/(# atoms the atom forms bonds with)</p>
<p>And for nonexistent’s question on calcium carbonate and nitric acid, why can’t it be:</p>
<p>H+ + CaCO3 –> HCO3- + Ca2+?</p>
<p>according to what my brain says, nitric acid is a very strong acid so it will dissociate H+ and NO3- to completion, so the carbonate ion from CaCO3 will have all H+ that it can bond with? Hence 2H+ + CO3 2- –> [H2CO3] –> H2O + CO2.</p>
<p>Kinetics:</p>
<p>I understand that if a reactant is quadrupled in amount formed while the other reactants stays constant, the order is 2. What if the reactant is tripled? What would be the order?
2^x = 3?</p>
<p>Is the rate of disappearance negative?</p>
<p>Also, all the questions I’ve dealt with have one reactant constant so you can determine the other. What if there is a case where the other reactant is NOT constant? I know there is a way to solve it, I just don’t know what it is. Any help is greatly appreciated. Thanks.</p>
<p>@ocean- also because calcium carbonate does not exist as CaCO3 in solution, it is soluble so it must as CO32- ions and Ca2+ ions. So H+ reacts with the carbonate ion</p>
<p>@aryus- I dont think that would be on the exam because that is not a whole number for the order of the reactant. Usually collegeboard only asks whole numbered orders.</p>
<p>also, if more than one reactant is changed. first you would find the order of one of those reactants (from looking at other experiments), then you would see what that reactant would do to the rate. from this, you can calculate the order of the other reactant from doing the same method as if only one reactant changed.</p>
<p>For example, if we had two chemicals, [O2] and [N2O5], lets say they start out at 0.1M each and the rate is 1M/sec. From looking at other experiments you can easily figure out the order of at least reactant since they always give one experiment where all of the other concentrations stay constant and only changes. So lets say that you find the order of the [O2] to be 1. And now we need to find the order of [N2O5], and lets say in experiment 3: [O2] = 0.2 and [N2O5] = 0.2 and the rate was found to be 8M/sec. we now have a problem since both of the concentrations changed, but we do know that the order of O2 is 1. So this means that when ever you double the concentration of O2 the rate should be doubled. So lets remove the effect of O2 from the rate, so the rate would be 4M/sec if [O2] was just 0.1 Now we can easily find out the order of N2O5 since the [O2] is held constant from the first experiment to the third. Which is 2^x = 4 and we find that x = 2. So the order of O2 is 1 and the order of N2O5 is 2.</p>
<p>There might be a formula for this, but this how I always think about. Sorry if its kind of hard to understand.</p>
<p>Here is a site that may help with Kinetics: [Introducing</a> Chemical Kinetics - Rate Equations](<a href=“http://www.avogadro.co.uk/kinetics/rate_equation.htm]Introducing”>http://www.avogadro.co.uk/kinetics/rate_equation.htm) </p>
<p>So basically, if the concentration of one reactant is kept constant while the other concentration doubles…
Rate stays the same = Zero order (w/ respect to that reactant)–2^0
Rate doubles = 1st order (" " “)–2^1
Rate quadruples = 2nd order (” " “)–2^2
Rate octuples = 3rd order (” " ")–2^3</p>
<p>If the concentration of one reactant is kept constant while the other concentration triples…
Rate stays the same = Zero order (w/ respect to that reactant)–3^0
Rate triples = 1st order (" " “)–3^1
Rate x9 = 2nd order (” " “)–3^2
Rate x 27 = 3rd order (” " ")–3^3</p>
<p>Now, I did come across a problem once where one reactant’s concentration was not kept constant when the other one was. Here is what I did, but please note that this may NOT be foolproof, it worked for this problem but may not for others:</p>
<p>Experiment 1: Initial [NO] = 0.0160; Initial [Br2] = 0.0120; Rate = .000324
Experiment 3: Initial [NO] = 0.0320; Initial [Br2] = 0.0060; Rate = .000642</p>
<p>I want to find the order of [NO], wwhose conc. is doubled, but [Br2] is halved. Thus, I adjusted the conc. of [Br2] in Experiment 3 by simply multiplying it by 2. However, with the algebraic rule in mind that states you must do the same thing to everything, i multiplied the rate in Experiment 3 by 2. Keep [NO] in Exp. 3 the same even though this may seem contradictory.</p>
<p>So now the new data read as follows:</p>
<p>Experiment 1: Initial [NO] = 0.0160; Initial [Br2] = 0.0120; Rate = .000324
Experiment 3: Initial [NO] = 0.0320; Initial [Br2] = 0.0120; Rate = .001284</p>
<p>So now .001284 / .000324 = 4, thus making the reaction 2nd order with respect to [NO].</p>
<p>rate is easy. I have been fascinated though, of how I could derive the equation myself using calculus.
-dA/dt = k<a href=“try%20sub’bing%20N%20too.%20this%20should%20prove%20that%20half-lives%20from%20nuke%20decay%20are%20of%20first%20order%20kinetics”>A</a>
-dA/dt = k[A]^2
-dA/dt = k[A]^3
-dA/dt = k[A]^4…
For calculus students…
try this. the results are revealing.
very interesting… or is it just for me?</p>
<p>^ wow crazy</p>
<p>also for calculus students</p>
<p>1st order integrated rate law: ln[A] - ln[A] intial = -kt</p>
<p>2nd order integrated rate law: 1/[A] - 1/[A] intial = -kt</p>
<p>Goodbyehello//I think i equated rate of disappearance and the actual rate, no?
and for clarification, for the diffy equations they are separable and use tf for upper limit and t0 for lower limit</p>
<p>Alright I have a really quick question. Can anybody explain why (b) is the answer:</p>
<p>Which of the following aqueous solutions has the highest boiling pt at 1.0 ATM</p>
<p>a) .20 M CaCl2
b) .25 M Na2SO4
c) .30 M NaCl
d) .30 M KBr
e) .40 M C6H12O6</p>
<p>kmi = change in freezing + boiling </p>
<p>Na2SO4 - 3 ions (2 Na and 1 SO4) *.25 m * k = highest value of all</p>
<p>ok this is just the use of colligative properties. delta T is kmi and if you have a species that is high in BOTH the van’t hoff factor and molality, then that species has the highest boiling point elevation, hence the highest boiling point.
let’s put them in a competition to see if who wins 
A. i = 3. m = 0.60 PHAIL
B. i = 3. m = 0.75 WIN
C. i = 2 m = 0.60 PHAIL
D. i = 2 m = 0.60 PHAIL
E i = 1 m = 0.40 BIG EPIC PHAIL</p>
<p>Dude so helpful thanks guys! Basically I think I’m screwed for questions that test the properties of compounds though because I keep getting those wrong on the pratice mcs I’ve taken so far</p>
<p>This is extremely basic but i’m having a little bit of trouble calculating ox. # of compounds. I get all the easy ones where you have to set the given comp. equal to zero, but the polyatomics i’m always off on when I try to follow the rule “set charges and x equal to charge of poly ion”</p>
<p>Here’s a problem:</p>
<p>What is ox. # of non-oxygen element in the poly ion of iron (III) nitrate</p>
<p>my calculations: + 3 + (-6) + x = -1…x = +2</p>
<p>Do I just disregard charge of iron? if so, why? the answer is slated to be +5 for N (x)</p>
<p>Another one:</p>
<p>What is ox. # of non-oxygen element in the poly ion of sodium hydrogen phosphate</p>
<p>formula: Na2HPO4</p>
<p>+2 (Na) + 1 (H) + (-8) (O) + x = -1…x = +4</p>
<p>The supposed answer is +5 as the ox. # for P</p>
<p>I probably just need a straightening-up on rules for ox. # calc. w/ polyatomics. Thanks.</p>
<p>How can I limit stupid mistakes? I just took the 2002 released exam, and I got 39 right, 24 wrong, and I skipped 12. Out of those 24 that I got wrong, I should have easily gotten 8 right. Am I doomed to make stupid mistakes on test day too? 
This puts me at a 3…</p>
<p>Anyone know the minimum points out of 150 required to get a 5?</p>
<p>ive heard you need somewhere in the high 90s for a 5</p>
<p>^Really?
I’ve always heard it’s something along the lines of 70-75% correct= 5.</p>