ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: 2NE1 on April 13, 2014, 01:34:46 pm
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According to researchers, a human being exhales around 850 g of carbon dioxide per day and, on average,
a healthy deciduous tree absorbs and stores 22 kg of carbon dioxide (as wood) annually. 6 moles CO2(g) is
fixed into 180 g of wood
Using the information above and assuming a year is 365.25 days, how many of these deciduous trees does a
human roughly need in order to have a carbon neutral life (just from breathing) if the average lifespan of a
human is 85 years?
A. 47
B. 818
C. 4907
D. 35 985
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I think the ambiguous wording of this question is what makes it difficult. I've managed to get an answer that isn't listed so probably interpreted it the wrong way.
My logic.
A human will exhale 850*265.25 grams of CO2 a year. One year of human = 310462.5 grams CO2.
If a single tree can absorb 22000 grams of CO2 a year, then we simply need 310462.5/22000 trees to be carbon neutral?
So we need 14 trees at any given time to be carbon neutral.
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I did it the same way as ^^^
If they want an answer more complicated they need to provide more information like the weight of the trees or something. The 6 mols per 180g is not a useful piece of info on its own. Also do you take into account the mass of the tree as it grows, how long the average tree life span is, etc.
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According to researchers, a human being exhales around 850 g of carbon dioxide per day and, on average,
a healthy deciduous tree absorbs and stores 22 kg of carbon dioxide (as wood) annually. 6 moles CO2(g) is
fixed into 180 g of wood
Using the information above and assuming a year is 365.25 days, how many of these deciduous trees does a
human roughly need in order to have a carbon neutral life (just from breathing) if the average lifespan of a
human is 85 years?
A. 47
B. 818
C. 4907
D. 35 985
I got B.
n=m/M to find out how many grams of CO2 is stored in 180 grams of wood.
264 grams of CO2 per 180g of wood.
22kgs of wood in a tree, so 22/0.18= 32.266667 kgs of CO2
human produces 0.850 kgs a day, 0.850 x 365.25= 310.4625 kg of CO2 a year.
310.4625 x 85 years = 26,3893.3125 kgs of CO2.
Divide by 32.26667 = ~818
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22kgs of wood in a tree, so 22/0.18= 32.266667 kgs of CO2
I understand you are probably correct, but the question states that there is 22kg of CO2 per tree, not that a tree weighs 22kg.
tree absorbs and stores 22 kg of carbon dioxide
is the part that is wrong I believe, and should be reworded.
Bad question.
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I think the ambiguous wording of this question is what makes it difficult. I've managed to get an answer that isn't listed so probably interpreted it the wrong way.
My logic.
A human will exhale 850*265.25 grams of CO2 a year. One year of human = 310462.5 grams CO2.
If a single tree can absorb 22000 grams of CO2 a year, then we simply need 310462.5/22000 trees to be carbon neutral?
So we need 14 trees at any given time to be carbon neutral.
Thats what I got too!! What I don't get is the 6 moles per 180g bit, it must have something to do with the working or the question has left bits out?
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Check out swagsxcboi's post, I believe that's the 'correct' way of answering it.
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Check out swagsxcboi's post, I believe that's the 'correct' way of answering it.
Yes, I tried doing it that way. Thanks for everyones help!
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Also another question which has a high degree of ambiguity (which is the magic acid)
Question 6
In 1994, George A Olah, a Hungarian-born American scientist, won the Nobel Prize in Chemistry for his
work on superacids. These are acids that are a billion billion times stronger than 100% sulfuric acid.
These superacids can be used to liquefy coal and to ‘crack’ heavy mineral oils, which could help improve
the processes involved in making petrochemicals. One such superacid is ‘magic acid’ and it is made by
mixing together antimony pentafluoride (SbF5) and fluorosulfonic acid (HSO3F). The reaction is
shown below.
2HSO3F(aq) + SbF5(aq) → FSO3SbF5(aq) + H2SO3F+
(aq)
What concentration, in mol L–1, of magic acid is formed when 2.0 M solutions of each reactant, both with a
volume of 250 mL each, are mixed together? (Assume the reaction goes to completion.)
A. 2 × 10–2
B. 0.25
C. 1.0
D. 2.0
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Where'd you get the questions 2NE1? I'm just curious (:
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Where'd you get the questions 2NE1? I'm just curious (:
I actually have no idea where they are from, it was on a link on our student chemistry database.
Do you have any idea how to get an answer for the other question? Because I think its 0.5 but thats not one of the options
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I actually have no idea where they are from, it was on a link on our student chemistry database.
Do you have any idea how to get an answer for the other question? Because I think its 0.5 but thats not one of the options
H2SO3F+ is the magic acid.
n(HSO3F)=2*0.25=0.5mol
n(H2SO3F)/n(HSO3F)=1/2
Hence, n(H2SO3F)=1/2*0.5=0.25mol
C(H2SO3F+)=n(H2SO3F+)/V(H2SO3F+)=0.25/0.25=1 mol L-1
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H2SO3F+ is the magic acid.
n(HSO3F)=2*0.25=0.5mol
n(H2SO3F)/n(HSO3F)=1/2
Hence, n(H2SO3F)=1/2*0.5=0.25mol
C(H2SO3F+)=n(H2SO3F+)/V(H2SO3F+)=0.25/0.25=1 mol L-1
why did you not divide the moles by 0.500 but 0.250 to work out the concentration?
because isn't the total volume 500mL as there are 250mL of each reactant?
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why did you not divide the moles by 0.500 but 0.250 to work out the concentration?
because isn't the total volume 500mL as there are 250mL of each reactant?
Are you referring to the last line of the working out? To find the concentration of a particular reactant, we must divide the number of moles of that reactant by the volume of that reactant. In this case, we calculated H2SO3F+ to have 0.25 moles. There can't be 500ml of magic acid alone as there is another product too! Yes the total volume is 500, but to find the volume of magic acid, we must halve this to get 250ml or 0.25L.