In an experiment 0.10g of gas found to occupy measured at standard pressure(1.05*205pa) and27 degree.The relative molecular mass is?
Answers
Answer:
To measure the rate of organismal respiration, a 10mL microrespirometer was constructed. These measure relative volume as oxygen is consumed by respiring organisms.
However, as oxygen gas is consumed during aerobic respiration, it is replaced by CO
2
gas at a ratio of one molecule of CO
2
for each molecule of O
2
. However, in the following protocol, the CO
2
produced is removed by a small piece of cotton treated with potassium hydroxide ( KOH ).
KOH reacts with CO
2
to form the solid potassium carbonate ( K
2
CO
3
). 5mL of germinating pea seeds were placed in one setup while a control microrespirometer was filled with 5mL of glass beads in another.
Over the course of 20 minutes, changes in the volume were measured by following the distance of a red soap bubble placed on the end of a 40uL capillary tube at time =0.
In one alternate experiment, instead of using germinating peas, 5 mL of Drosophila melanogaster were placed in the microrespirometer.
What would this change have on the results?
Explanation:
No change in consumption of O
2
will occur, under this experimental condition as there is no mention of increased activity by Drosophila. In normal conditions as the chemical process of breakdown of Glucose is same to release ATP using O
2
in both plants and animals the rate of consumption of O
2
will be same.
Germinating seed are below the surface of the soil, so they do not photosynthesize, hence no release of O
2
as a byproduct of the process which would otherwise be utilized for respiration by the plant.
So, the correct answer is 'No change in oxygen consumption would occur because they respire at the same rate'