lydiafoof
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#1
Report Thread starter 6 years ago
#1
The student’s results are shown in Table 3.
Table 3

Time / minutes Total volume of gascollected / cm3
10 0.3

20 0.9

30 1.9

40 3.1

50 5.0

60 5.2


Calculate the rate of gas production in cm3 g–1 min–1 during the first 40 minutes of thisinvestigation. Show your working.

Can anyone help me with this question I have tried and have got the wrong answer.

The markscheme is:

0.0155 / 0.016 = 2 marks;
0.0775 / 0.077 / 0.078 / 0.08 = 1 mark
/ 0.62 = 1 mark

Thank you
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sidney1
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#2
Report 6 years ago
#2
http://www.thestudentroom.co.uk/show...1#post56197061
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lydiafoof
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#3
Report Thread starter 6 years ago
#3
Thank you!
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Aisha Catt
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#4
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#4
Ok so it tells you in cm^3 g^-1 min^-1 which basically gives you the formula cm3/(gxmin)So at 40 min 3.1/(5x40) = 0.0155Hope that helps, its just the way I think about it.
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JHTY
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#5
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#5
(Original post by Aisha Catt)
Ok so it tells you in cm^3 g^-1 min^-1 which basically gives you the formula cm3/(gxmin)So at 40 min 3.1/(5x40) = 0.0155Hope that helps, its just the way I think about it.
Where did the 5 come from when you did 5x40?
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emilyyy6
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#6
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#6
(Original post by JHTY)
Where did the 5 come from when you did 5x40?
because in the question it says that 5g of yeast is used
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