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How would you go about finding the density or air? I think you have to find the mass && volume of air first, but i dont know how to do that either ... helppp???

2006-09-11 09:41:06 · 4 answers · asked by jayyyy 1 in Education & Reference Homework Help

4 answers

The volume would be easy to measure. Under normal pressure, whatever is in the given container is the volume.

The mass would be a bit difficult to measure, as we have the same gas with same density all around it. It will simply be "weightless" if we try to measure air in air.

One way I can think of doing is, take a volume of air and compress it to let's say 10th of the volume. Then, the same amount of the air, now compressed will still contain the same amount but now is heavier than the air around it, allowing you to measure.

I think, you have to compensate for the weight of the air around it by adding 10%. (reciprocal of *10 compression)

2006-09-11 09:50:51 · answer #1 · answered by tkquestion 7 · 0 1

Yes, find the mass of the components first.

You'd have to think what gases constitute air. How much of each gas (in percentage) is present. You should look that up, but from the top of my head - Nitrogen is about 70 %, Oxygen is about 20%, You can get the densities for those from the periodic table.

2006-09-11 09:45:54 · answer #2 · answered by Snowflake 7 · 0 0

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Molecular weight of air = (80*28) + (20*32) / 100 = 28.8 We take 80 % of N2 and 20 % of O2. (Not correct but very near). One mole of a gas = 22.4 dm³ (OK, at O°C and 1 atm). So the density of air is 28,8 / 22,4 = 1.28 g / dcm³ = 0.00128 g / cm³. As your student found 0.1 g for 100 cm³, the answer is of course A. In reality it also is A, which is the nearest to the theory. We have calculated it but not really in a correct way. Remember it is not exactly 80-20 and is it at O°C and 1 atm ?

2016-04-03 01:43:23 · answer #3 · answered by Anonymous · 0 0

you would need very sensitive equipment to accurately "mass" the air. volume will be easy to measure.

2006-09-11 09:45:35 · answer #4 · answered by SwampDog 2 · 0 0

Try this website for starters.

2006-09-11 09:47:15 · answer #5 · answered by Peter R 2 · 0 0

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