<p>how do you know if something is first order or second order?</p>
<p>@MasterMaestro, Me2 plz!</p>
<p>Ok, mathpop: E = hc/lambda</p>
<p>Master: We need to know general procedure for stuff like standardization, dilution, caliormetry, titration, gravimetric analysis, etc. </p>
<p>We need to be able to identify sources of errors and how they effect the experiment. We must be familiar with the names of equipment, their proper uses, and safety procedures. We also need to know stuff on sig digs and precision.</p>
<p>I found this helped me understand reaction orders: <a href=“http://www.chem.ufl.edu/~itl/4411/lectures/lec_l.html”>http://www.chem.ufl.edu/~itl/4411/lectures/lec_l.html</a></p>
<p>@zah0813 to find the order or a reaction you need to first create the rate law: Rate = k[X][Y]. Once you have that equation just count all of the exponents that is in the equation. So the same rate law that I gave you the order would be two (or 2nd order) because the exponents of the concentrations of X and Y are 1 and then I add those up and it would equal 2. Another example would be: Rate = k[X]^2[Y]. So for this case the order would be 3 because you add up all of the exponents. I hope that made sense</p>
<p>I also was wondering if this test is going to be harder or easier due to the changes. I took the only official practice test that had the changes applied to the test, and it was CRAZY HARD! So I’m hoping that it could be toned down a bit because other practice tests haven’t been nearly as hard as that new test. Or that might just be me.</p>
<p>Also, for radioactive decay, the rate is always first order. Then you can calculate half-life and some other things.</p>
<p>If they give you a table, you must keep one concentration constant and like double the other one to see how it effects the overall rate. If doubling one concentration doubles the rate, the exponent is 1 FOR THAT ONE CONCENTRATION. If it grows by 4, then the exponent is 2. That’s a VERY common question for FRQ and MC.</p>
<p>An example of the table thing I’m talking about is on the link near the top that coolschool posted.</p>
<p>xbox did you get unbanned yet…</p>
<p>@kingofxbox99 can you explain to me why they are both positive in terms of what spontaneity is depending on? </p>
<p>Nope, you?</p>
<p>Can someone explain 57 on the practice test for me? It looks to me that they all produce a straight line so I don’t understand why it’s first order</p>
<p>If someone takes a picture of the practice test and posts it, I’ll help everyone here. </p>
<p>do you guys have a concise, good source that explains acidbase equilibrium in really simple terms? It’s the most confusing thing for me that we’ve learned all year.</p>
<p>@KiterPatel Sure!</p>
<p>So if a process is spontaneous, delta G must be negative. From our formula sheet, we know that</p>
<p>G = H - TS</p>
<p>So if at a low temperature, its not spontaneous, so G is positive. Lets pretend T is 1K (very cold).</p>
<p>G = H - 1*S = H - S If G is positive, then H must be larger (and positive) than S.</p>
<p>For high temperatures, (lets assume T = 1000K), it IS spontaneous so G is negative.</p>
<p>G = H - 1000S. For G to be negative, S must be positive and H can be either (since 1000S is WAY bigger than H).</p>
<p>Combining the two, we find that S must be positive and H must be positive and larger than S.</p>
<p>Does that make sense?</p>
<p>In the first column, the pressure is being halved each time, which shows you that it is a half-life decay which means that it is first order. the ln(PN2O5) column is decreasing by exactly the same number each time, so that one is perfectly linear. ln§ vs. time is first order</p>
<p>KiterPatel, I posted the answer on the other page. :)</p>
<p>@kingofxbox99 yes thank you!</p>
<p>bump: “do you guys have a concise, good source that explains acidbase equilibrium in really simple terms? It’s the most confusing thing for me that we’ve learned all year.”</p>