allthebest Posted October 4, 2016 Report Share Posted October 4, 2016 (edited) I have some questions from paper 1. 1. the first one is about first order reaction. I thought the answer was D, but the mark scheme says it is C. I thought C was for second order reactions... anyone can explain why the answer is C? 2. I don't quite understand conjugate acid-base pairs. I thought they have difference of 1 H atom. The answer is D, everything's correct. Please help by explaining... 3. I thought those results are precise, close to each other, and accurate, close to actual value... isn't it? which value should we consider when determining precision and accuracy? answer is B 4. I have no idea how to get the answer for this one. What are the key relationships here? answer is D Thank you in advance Edited October 4, 2016 by allthebest Reply Link to post Share on other sites More sharing options...
kw0573 Posted October 4, 2016 Report Share Posted October 4, 2016 20) Maybe knowing derivatives help with this question. Zeroth order is constant rate, which means Y decreases steadily over time. First order is rate is proportional to the concentration of Y, which means as Y decreases, so does the change in Y. Specifically, if there is half of Y left, the rate or derivative is half as much. There is nothing to do with integrated rate laws if that's your concern. Simply you compare the slopes at Y = 1 and Y = 0.5. 24) Specifically, Bronsted-Lowry acid-base pairs differ by a charge and an H. In fact, all acid base pairs provided do indeed differ by a charge and an H. The acid has one more H than the base (Bronsted-Lowry definition of acids/bases). Also, the acid as one less negative charge (or one more positive charge) than the base (Lewis definition of acids/bases). Reply Link to post Share on other sites More sharing options...
allthebest Posted October 4, 2016 Author Report Share Posted October 4, 2016 (edited) 10 minutes ago, kw0573 said: 20) Maybe knowing derivatives help with this question. Zeroth order is constant rate, which means Y decreases steadily over time. First order is rate is proportional to the concentration of Y, which means as Y decreases, so does the change in Y. Specifically, if there is half of Y left, the rate or derivative is half as much. There is nothing to do with integrated rate laws if that's your concern. Simply you compare the slopes at Y = 1 and Y = 0.5. 24) Specifically, Bronsted-Lowry acid-base pairs differ by a charge and an H. In fact, all acid base pairs provided do indeed differ by a charge and an H. The acid has one more H than the base (Bronsted-Lowry definition of acids/bases). Also, the acid as one less negative charge (or one more positive charge) than the base (Lewis definition of acids/bases). I don't still quite get no.20...but you're sure the answer is right, right? how should I compare the slope? and if you would please answer the other 2 question I edited below, please. Thank you Edited October 4, 2016 by allthebest Reply Link to post Share on other sites More sharing options...
kw0573 Posted October 4, 2016 Report Share Posted October 4, 2016 Precise = lots of significant digits. Or alternatively, precise means the measurement is repeatable (close to each other, like you said). Yet when the precision has less significant digits, it is not precise enough. I would say this is accurate since 2.3 and 2.4 is not that far. IB may disagree. I would say D. not precise, accurate. Reply Link to post Share on other sites More sharing options...
kw0573 Posted October 4, 2016 Report Share Posted October 4, 2016 6 minutes ago, allthebest said: I don't still quite get no.20...but you're sure the answer is right, right? how should I compare the slope? and if you would please answer the other 2 question I edited below, please. Thank you Hi. Perhaps you should not edit your responses so frequently? Submit all questions in one go. In D, it's more plausible that the green line has half the slope as red line, but in C they are too close. d[Y]/dt = rate (which is proportional to Y). so compared to Y = 1, Y = 0.5 has half the slope. 2. PV = nRT, n = PV/(RT) 27°C = 300K, 372°C = 600 K Therefore at double temperature and double pressure, n remain the same ==> D. Reply Link to post Share on other sites More sharing options...
allthebest Posted October 4, 2016 Author Report Share Posted October 4, 2016 11 minutes ago, kw0573 said: Hi. Perhaps you should not edit your responses so frequently? Submit all questions in one go. In D, it's more plausible that the green line has half the slope as red line, but in C they are too close. d[Y]/dt = rate (which is proportional to Y). so compared to Y = 1, Y = 0.5 has half the slope. 2. PV = nRT, n = PV/(RT) 27°C = 300K, 372°C = 600 K Therefore at double temperature and double pressure, n remain the same ==> D. so for no.20, you're saying the answer is D, aren't you? Reply Link to post Share on other sites More sharing options...
kw0573 Posted October 4, 2016 Report Share Posted October 4, 2016 4 minutes ago, allthebest said: so for no.20, you're saying the answer is D, aren't you? I was. But I was also mistaken. It's asking about half-life. Very tricky. It wants Y = 0.5 compared to Y = 0.25 Reply Link to post Share on other sites More sharing options...
allthebest Posted October 4, 2016 Author Report Share Posted October 4, 2016 1 minute ago, kw0573 said: I was. But I was also mistaken. It's asking about half-life. Very tricky. It wants Y = 0.5 compared to Y = 0.25 Oh my. if its half life, now I get it. Thanks a lot!! Reply Link to post Share on other sites More sharing options...
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