<p>@kingofxbox99 In the problem it says it is a second order reaction</p>
<p>@NewJersey25 What? I see it and it says first order reaction. </p>
<p>@hawkace @kingofxbox99
Found the issue, the link shuffles the problems everytime you click it. Ill post the question here. </p>
<p>For a second order reaction with a rate constant k = 3.13 x 102 L mol-1 s-1, how long does it take for the concentration of the only reactant to become 70% of the original amount?
A) 0.00137 seconds
B) 0.0761 seconds
C) 0.912 seconds
D) 0.00114 seconds
E) 0.00746 seconds</p>
<p>Can someone explain to me how to do the First order and second order problems? How can you tell if its a first, second, or zero order just be looking at the data chart?</p>
<p>For example, #57 on the practice exam</p>
<p>@kbbgrizzly you have to plot the data in three graphs. the first being [A] vs. time, the second is ln[A] vs. time and the third is 1/ [A] vs. time</p>
<p>if the first graph produces a straight line you have a zero order rxn
if the second graph produces a straight line you have a 1st order rxn
and the third graph is a 2nd order rxn</p>
<p>ohhhhhhh
Thanks @NewJersey25 :)</p>
<p>I wish I could stop time, study all I want and turn time back on to take the exam…lol</p>
<p>@coolschool Hi one more question–I got accepted to CSULB and I was wondering if taking the test and getting a 1 better than not taking it at all? Or is it better to not take it all and not fail it? Because even though I’m online schooling–we really are independent studying and it’s been really hard. I took the Princeton Review 2013 Practice Exam and got a 25/75. I’m really nervous and I haven’t a clue what to do for the free response questions.</p>
<p>@xxx123456xxx I heard that it’s better to take the exam although you fail it. It shows colleges that you can follow through the AP course just by taking the exam.</p>
<p>@xxx123456xxx Do you get credit for this AP exam?</p>
<p>In my personal opinion, taking the test and getting a 1 is better. First off, you don’t HAVE to send AP scores to universities if you don’t want. Secondly, taking the AP exam prepares you for university and gives you a glimpse of good problems and critical thinking skills. </p>
<p>That’s just my two cents.</p>
<p>Can anyone answer this? </p>
<p>For a second order reaction with a rate constant k = 3.13 x 102 L mol-1 s-1, how long does it take for the concentration of the only reactant to become 70% of the original amount?
0.00137 seconds
0.0761 seconds
0.912 seconds
0.00114 seconds
0.00746 seconds</p>
<p>CSULB requires a 3 on the exam to get 6 credits in Chemistry, so I would say that it’s worth a shot to take the exam since it’ll ultimately save you a lot more money. </p>
<p>@HawkAce I do get credit but I doubt I’ll get a 3 I think it’ll be a miracle to get a 2 honestly. I heard the new test for this year was hard too and I just don’t know. @kingofxbox99 yeah that’s what I was planning on. I was thinking of rush delivering it, after I see my scores, if I got a 3 or higher. I’m taking 4 AP Exams and I know I’ll fail my Bio one too. </p>
<p>Im pretty sure its .00137 seconds
(1/.7)-(1/1)= kt
since k= 313 all you do is solve for time </p>
<p>@KbbGrizzly I know–I’m just really scared like if I don’t know the stuff in the FRQ can I just leave it blank? Because I really don’t know. Standardized tests terrify me.</p>
<p>For some reason when I clicked .00137 it said it was incorrect which is why I was confused</p>
<p>@xxx123456xxx you can always get partial points for the FRQ section. I feel you since I’m dying from the FRQs right now hahah, but if you can study hard today and relax and do as much as you can with your best ability tomorrow then you’ll be fine ^^</p>
<p>If anyone could explain the significant figures part of the lab question, it would be really helpful</p>
<p>@NewJersey25 I found a similar question on this site #15
<a href=“http://www.dublinschools.net/Downloads/Sciencegeek%2012.pdf”>http://www.dublinschools.net/Downloads/Sciencegeek%2012.pdf</a>
hope this helps 
but the equation is different…the question you asked is second order</p>