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Chemical equilibrium in oxyhydrogen gas sample?
Hey, My question would be, why after the Knallgasse reaction under normal conditions, H2O does not react back into its reactants? My guess would be that water is a stable molecule due to the hydrogen bonds, and therefore it would require a lot of energy to overcome them. But how could this be explained by…
The estimated total volume of all oceans should be easy to find out.
Convert quantity of substance to number of molecules and divide by volume.
So the volume is: 1.33*10^9 km^3
And the 1.33*…. I have to multiply 6*10^23 and then share again by 1.33*… or?
1 mol = 6.02*10^23 particles. 1/4 mol of particles are available. How many particles are they?
This is distributed over 1.335*10^9 cubic kilometers.
Conversion to cubic meters:
1,335*10^9 cubic kilometers = 1,335 * 10^9 * (10^3 m)^3
= 1,335 * 10^9 * 10^9 m^3
Herexception is 1 liter. This has to be converted into liters (1 l = 1 dm^3, 1 dm = 1/10 m):
… = 1.335 * 10^9 * 10^9 * (10 dm)^3
= 1,335 * 10^9 * 10^3
Now I’m confused. Can you explain that to me a little bit easier?
Note the units: The number of molecules in one liter, i.e. a dm3 (0.1 m)^3 ), the volume is given in km3 (1000 m)^3 ).
Also note what the Avogadro number must be multiplied by.
Somewhere you need to get the water volume of all world seas, convert this volume into liters and share the number of alcohol molecules that correspond to 0.25 mol by the number of litres of water in the world seas.