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    (ii) At 30 °C, one student obtained the following results.
    Calculate the mean rate of gas production. Give your answer in cm3


    Answer ............................... cm3

    (2)
    Volume of gas
    collected in 5 minutes /
    cm3
    Result 1 Result 2 Result 3
    38.3 27.6 29.4

    (iii) If aerobic respiration had been investigated rather than anaerobic respiration, how
    would you expect the volumes of gas collected at 30°C to differ from these results?
    Explain your answer.


    Anyone help explain this question to me?
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    Hi,

    I would be happy to help you and in fact walk you through the logic and how to work it out, but I need to see the whole Q. There is no copy of the description of the experiment, which gas is being collected (CO2??), or what the first part of Q asked. Please provide ALL these.

    M
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    (Original post by kathy9)
    (ii) At 30 °C, one student obtained the following results.
    Calculate the mean rate of gas production. Give your answer in cm3


    Answer ............................... cm3

    (2)
    Volume of gas
    collected in 5 minutes /
    cm3
    Result 1 Result 2 Result 3
    38.3 27.6 29.4

    (iii) If aerobic respiration had been investigated rather than anaerobic respiration, how
    would you expect the volumes of gas collected at 30°C to differ from these results?
    Explain your answer.


    Anyone help explain this question to me?
    1. When it says rate of gas production, in my mind, i think you should find out how much volume of gas is produced per minute or second, but you've only mention cm3 as the unit for the answer so its just adding results 1,2 and 3 and dividing by 3. If however youve simply forgotten the units then once you get the mean volume of gase yuo'd have you divide it by 5 to get how much cm3 of gas is produced per minute, but if the unit of time is seconds youd just divide that value by 60 to get cm3 per second.

    2. Im going to assume that the gas we're talking about is CO2 and its being produced by an organism. If aerobic resp was investigatedf i reckon that more gas would be produced, since it goes through more processes than anaerobic resp that produces CO2 as a waste product, such as the link reaction(decarboxylation of pyruvate into acetate) and then the kreb cycle (more decarboxylation occurs i.e. release of CO2 from products like the 6 C compound that converts into the 4 C compound which would react with acetyl coA and restart the whole cyle). And these processes do not occur in Anaerobic respiration since it only involves glycolysis so no where near as much CO2 is produced in anaerobic resp than in aerobic.

    Hope this helps, if you have the markscheme id appreciate it if you could let me know if i missed anything or if i was wrong XD
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    (Original post by macpatelgh)
    Hi,

    I would be happy to help you and in fact walk you through the logic and how to work it out, but I need to see the whole Q. There is no copy of the description of the experiment, which gas is being collected (CO2??), or what the first part of Q asked. Please provide ALL these.

    M
    Hi the whole question is
    Some students investigated the effect of temperature on the rate of anaerobic respiration in yeast. The apparatus they used is shown in the diagram. The yeast suspension was mixed with glucose solution and the volume of gas collected in five minutes was recorded. (i) Each student repeated the experiment and the results were pooled. Explain the advantages of collecting a large number of results. ................................ ................................ ................................ ............. ................................ ................................ ................................ ............. ................................ ................................ ................................ ............. ................................ ................................ ................................ ............. (2)


    (ii) At 30 °C, one student obtained the following results. Calculate the mean rate of gas production. Give your answer in cm3 .
    Answer ............................... cm3

    (2) Volume of gas collected in 5 minutes / cm3
    Result 1 Result 2 Result 3 38.3 27.6 29.4
    (iii) If aerobic respiration had been investigated rather than anaerobic respiration, how would you expect the volumes of gas collected at 30°C to differ from these results?

    I am only stuck with the last question. if this is unclear i got the question from the respiration pdf on the physics and maths tutor website
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    (Original post by Yme2day)
    1. When it says rate of gas production, in my mind, i think you should find out how much volume of gas is produced per minute or second, but you've only mention cm3 as the unit for the answer so its just adding results 1,2 and 3 and dividing by 3. If however youve simply forgotten the units then once you get the mean volume of gase yuo'd have you divide it by 5 to get how much cm3 of gas is produced per minute, but if the unit of time is seconds youd just divide that value by 60 to get cm3 per second.

    2. Im going to assume that the gas we're talking about is CO2 and its being produced by an organism. If aerobic resp was investigatedf i reckon that more gas would be produced, since it goes through more processes than anaerobic resp that produces CO2 as a waste product, such as the link reaction(decarboxylation of pyruvate into acetate) and then the kreb cycle (more decarboxylation occurs i.e. release of CO2 from products like the 6 C compound that converts into the 4 C compound which would react with acetyl coA and restart the whole cyle). And these processes do not occur in Anaerobic respiration since it only involves glycolysis so no where near as much CO2 is produced in anaerobic resp than in aerobic.

    Hope this helps, if you have the markscheme id appreciate it if you could let me know if i missed anything or if i was wrong XD
    the mark scheme is this:
    (iii) Volume(s) less/ no gas evolved; Glucose has RQ of 1.0; So (volume) CO2 evolved = (volume of) O2 taken in;
    Thanks for your answer, i understand it more than the markscheme lol. Still a bit confused
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    Hi, sorry about delay:-

    Let me guide you thru this Q as I promised:-

    The equation for Anaerobic respiration [with yeast - therefore really it is fermentation] is:

    C6H12O6 -----> 2C2H5OH + 2CO2

    So for each mole of glucose ANAEROBICALLY respired, 2 moles of CO2 released (No other gas involved going in or out - with me?) If not with me, shout out loud (NOT lol but sol ) -

    NOW let us proceed to AEROBIC respiration: (you are happy that this requires oxygen - yes............................. . Good!)

    Reaction:

    C6H12O6 + 6O2 -----> 6CO2 + 6H2O

    At 30C temp the water will remain liquid (boiling point of H2O = 100C)

    So 6 moles of oxygen going in AND how many moles of CO2 coming out....?...................

    Wait for it!............................
    ................
    Yes well done, 6 again.

    SO THERE IS NO NET change in any gas = no gas evolved - still with me or are you left a mile behind haha? I am probs 10 years older than you, but I am cycling faster! don't worry tell mum to google some recipes for hot chicken curry!

    OK therefore AS IN MARK SCHEME - no gas evolved (the mark scheme explains the meaning of Respiratory Quotient (RQ) which is 6/6 moles = 1.

    If you mention both these points you should be laughing [will get both marks] and you will not need any more tissues.

    ALSO DON'T rush to your local Tesco rt now to buy yeast - although as you see in the reaction that anaerobic respiration of yeast gives alcohol, you better wait till you are 18+ haha!

    Thanks, M.
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    (Original post by macpatelgh)
    Hi, sorry about delay:-

    Let me guide you thru this Q as I promised:-

    The equation for Anaerobic respiration [with yeast - therefore really it is fermentation] is:

    C6H12O6 -----> 2C2H5OH + 2CO2

    So for each mole of glucose ANAEROBICALLY respired, 2 moles of CO2 released (No other gas involved going in or out - with me?) If not with me, shout out loud (NOT lol but sol ) -


    NOW let us proceed to AEROBIC respiration: (you are happy that this requires oxygen - yes............................. . Good!)

    Reaction:

    C6H12O6 + 6O2 -----> 6CO2 + 6H2O

    At 30C temp the water will remain liquid (boiling point of H2O = 100C)

    So 6 moles of oxygen going in AND how many moles of CO2 coming out....?...................

    Wait for it!............................
    ................
    Yes well done, 6 again.

    SO THERE IS NO NET change in any gas = no gas evolved - still with me or are you left a mile behind haha? I am probs 10 years older than you, but I am cycling faster! don't worry tell mum to google some recipes for hot chicken curry!

    OK therefore AS IN MARK SCHEME - no gas evolved (the mark scheme explains the meaning of Respiratory Quotient (RQ) which is 6/6 moles = 1.

    If you mention both these points you should be laughing [will get both marks] and you will not need any more tissues.

    ALSO DON'T rush to your local Tesco rt now to buy yeast - although as you see in the reaction that anaerobic respiration of yeast gives alcohol, you better wait till you are 18+ haha!

    Thanks, M.
    Thank you so much! That was very helpful and funny cheers!
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    yes i agree with the above solutions
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    (Original post by macpatelgh)
    Hi, sorry about delay:-

    Let me guide you thru this Q as I promised:-

    The equation for Anaerobic respiration [with yeast - therefore really it is fermentation] is:

    C6H12O6 -----> 2C2H5OH + 2CO2

    So for each mole of glucose ANAEROBICALLY respired, 2 moles of CO2 released (No other gas involved going in or out - with me?) If not with me, shout out loud (NOT lol but sol ) -

    NOW let us proceed to AEROBIC respiration: (you are happy that this requires oxygen - yes............................. . Good!)

    Reaction:

    C6H12O6 + 6O2 -----> 6CO2 + 6H2O

    At 30C temp the water will remain liquid (boiling point of H2O = 100C)

    So 6 moles of oxygen going in AND how many moles of CO2 coming out....?...................

    Wait for it!............................
    ................
    Yes well done, 6 again.

    SO THERE IS NO NET change in any gas = no gas evolved - still with me or are you left a mile behind haha? I am probs 10 years older than you, but I am cycling faster! don't worry tell mum to google some recipes for hot chicken curry!

    OK therefore AS IN MARK SCHEME - no gas evolved (the mark scheme explains the meaning of Respiratory Quotient (RQ) which is 6/6 moles = 1.

    If you mention both these points you should be laughing [will get both marks] and you will not need any more tissues.

    ALSO DON'T rush to your local Tesco rt now to buy yeast - although as you see in the reaction that anaerobic respiration of yeast gives alcohol, you better wait till you are 18+ haha!

    Thanks, M.
    I suppose tahts a very succinct and easier way of explaingin it ahaha, was wondering though, my previous response...is it still scientfically valid? Like my rationale...does it make sense? What could i do to adopt the correct thinking, i.e. how can i come to the same conclusion as you did. Many thanks, and i loved your explanation, it was fantastic XD
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    (Original post by Yme2day)
    I suppose tahts a very succinct and easier way of explaingin it ahaha, was wondering though, my previous response...is it still scientfically valid? Like my rationale...does it make sense? What could i do to adopt the correct thinking, i.e. how can i come to the same conclusion as you did. Many thanks, and i loved your explanation, it was fantastic XD
    I think your logic and chain of thought was, in essence correct - in fact, in all honesty, I was also thinking of the TOTAL amount of CO2 being released being greater for aerobic respiration, which is absolutely correct as you said (AND you were right in stating the stages of aerobic respiration, too!); I think I got an unfair advantage in that I had sight of the whole Q, AND KNEW THAT THE EXPERIMENT DID NOT MEASURE ANY SPECIFIC GAS (e.g. CO2 using lime water), but the TOTAL mount of gas released, in effect NET AMOUNT!

    I don't think there is anything wrong with your rationale, and if I was marking your answer, I would give part credit for the detail of knowledge and logical analysis, even if you lose the mark for the actual answer being incorrect!

    M
 
 
 
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