Chemistry SL Paper 1 and Paper 2
Discussion forum for International Baccalaureate students.
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Re: Chemistry SL Paper 1 and Paper 2I'm not great at Chem, but I thought that parts of them were really quite tricky.(Original post by Cavid409)
What do you feel about it?
There were some weird questions on Paper 1, and I know I got 26 wrong (I put benzene ring and ketone, apparently it was benzene ring and ester). So not that great, but not a MAJOR catastrophe.
And I found Paper 2 hard, just because all the stuff I HATE came up, and none of the stuff I love (e.g. equilibrium). And I like acids and bases, but the Section B Q6 was mixed in with Enthalpy, which I hate! Plus, I screwed up the Iodine-131 and Carbon Dating questions.
Basically, not great. I'll have to abandon hope of getting a 7 and hope for a 6... -
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Re: Chemistry SL Paper 1 and Paper 2Haha, I know I got question 26 on Paper 1 wrong. I'm hoping i got about 26 right!(Original post by Cavid409)
You got 26 wrong? O.o I got all of them correct and had some difficulties in paper 2 so lost 6-7 points...
Losing 6-7 points is hardly catastrophic though! I know we aren't allowed to discuss Paper 3 yet, but you must be on track for a 7? I suppose you liked Paper 2 then...?
How do you know you got everything right on Paper 1 though? -
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Re: Chemistry SL Paper 1 and Paper 2Dedication!(Original post by Cavid409)
I checked everything with the syllabus... Every question..
I think I fell into the trap they set on q.1 as I put 6.02 x 10^22
Damnnn
Can you remember what you got for stuff? Oh, and what Section B question did you do? -
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Re: Chemistry SL Paper 1 and Paper 2Wasn't that Question 6? Acids and Bases/Enthalpy? Love Acids and Bases but I'm horrible at Enthalpy. And no equilibrium for the whole paper! I can kiss goodbye to 45...(Original post by Cavid409)
Yeah, of course i do... The answer is 6,62 x 10^23... Why? 0.1 (mol) x 11 (atoms) x 6,02 x 10^23 = 6,62 x 10^23... I chose the second question from section b... Ammonia and carbon dioxide..
Uhmmm what about drawing the tangent to the graph and finding the rate of decomposition for the first question? -
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Re: Chemistry SL Paper 1 and Paper 2I've been searching for a thread like this all over the internet. In fact, I just created this account so I could respond.
Paper 2 was far easier than Paper 1 (which I did pretty badly on)
I had doubts on the following questions; do you know what the right answer was?
• How many isomers of pentane there were
• concentration of NaOH after adding the water
• Given x and y, enthalpy of some N2H2 compound
• how accurate the measurement should be when measured ±.001 g
• moles of oxygen required for combustion of hydrocarbon CxHx+2 -
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Re: Chemistry SL Paper 1 and Paper 2I think i lost some points there)) in finding the rate i divided 4 by 120 and i don't is it was right or not... But are you sure that we have to draw a tangent to the curve?? Because i thought that the graph with catalyst will be steeper than the original one.. How did u write it?
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Re: Chemistry SL Paper 1 and Paper 2Are these questions from paper 2? Section b? Because i don't even remember these questions))(Original post by meep555)
I've been searching for a thread like this all over the internet. In fact, I just created this account so I could respond.
Paper 2 was far easier than Paper 1 (which I did pretty badly on)
I had doubts on the following questions; do you know what the right answer was?
• How many isomers of pentane there were
• concentration of NaOH after adding the water
• Given x and y, enthalpy of some N2H2 compound
• how accurate the measurement should be when measured ±.001 g
• moles of oxygen required for combustion of hydrocarbon CxHx+2 -
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Re: Chemistry SL Paper 1 and Paper 2they were paper 1 questions (sl)(Original post by Cavid409)
Are these questions from paper 2? Section b? Because i don't even remember these questions)) -
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Re: Chemistry SL Paper 1 and Paper 2Well, I just assumed that as it was 'rate', you had to find the gradient of the graph at 120 seconds (rate = gradient), so I drew a tangent and calculated (change in y / change in x). I can;'t quite remember what I got though...(Original post by Cavid409)
I think i lost some points there)) in finding the rate i divided 4 by 120 and i don't is it was right or not... But are you sure that we have to draw a tangent to the curve?? Because i thought that the graph with catalyst will be steeper than the original one.. How did u write it?
I did draw the curve higher than the original one though, as the reaction would have started off faster?
I think perhaps you are TZ1? We are TZ2...(Original post by meep555)
they were paper 1 questions (sl) -
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Re: Chemistry SL Paper 1 and Paper 2i din get the rate of decomposition crap but im pretty sure rate does not always equal to the gradient(Original post by beautiful-lion)
Well, I just assumed that as it was 'rate', you had to find the gradient of the graph at 120 seconds (rate = gradient), so I drew a tangent and calculated (change in y / change in x). I can;'t quite remember what I got though...
I did draw the curve higher than the original one though, as the reaction would have started off faster?
I think perhaps you are TZ1? We are TZ2...
if youve done physics then its like comparing to the equation:
y = mx
where m (gradient)= y/x
so the the gradient is the variable on the y axis divided by the variable on the x axis . And since time was not the variable on the axis, which im pretty sure, the gradient is not equal to the rate.
