how thick of a veneer is required if some of the water from the meteorite dissoc

Question

how thick of a veneer is required if some of the water from the meteorite dissociates and the resulting oxygen is used to oxidize one third of Fe+2 in the outer 800 km of the 6000 km proto earth to Fe3+? Assume that after dirrerentiation the outer 800 km of the proto-earth contains 7.0 wt.% Fe and has a density of 3.4g/cm^3.  Use these reactions

H2O -> H2 + 1/2O2

and

2FeO + 1/2O2 -> Fe2O3

Some hints that were given.... come up with amount of water and convert to moles.  (Answer should be 10-50km) . I need to oxidize, how much water is it, then add the water to the world oceans.difference of sphere volume. thanks.

Also given, oceans contain 1.4*10^24 g of water, and meteors are 15% water by weight.  

Will rate lifesaver and give all points.  Thank you!

Explanation / Answer

Surface area = S = 4*pi*R^2 = 4*3.14159*6000^2 km^2

S = 4.52x10^8 km^2

But 1 km^2 = 10^20 cm^2

S = 4.52x10^18 cm^2

The density of water is 1 gm/cm^3 so the volume occupied by 1.4x10^24 grams of water is simply the mass divided by the density.

V = 1.4x10^24 cm^3

S*D = V where D is the depth of water

D = V/S = 1.4x10^24 / 4.52x10^18 cm

D = 3.09x10^5 cm = 3.09 km

So if the water was spread evenly over the surface then the depth would be around 3.09 km

If you want to calculate the size of a single meteorite containing this much water then we can do that also.

M = mass of meteorite

0.15*M = mass of water = 1.4x10^24 grams ... 15% by mass

M = 9.334x10^24 grams

m = mass that is not water = 0.85*M = 7.933x10^24 grams

Volume of water = 1.4x10^24 cm^3 .... from before

Other volume = mass/density = 7.933x10^24/3 = 2.644x10^24 cm^3

Total volume = 4.044x10^24 cm^3

V = (4/3)pi*R^3 ... for a sphere

R = [(3/4)V/pi]^(1/3)

R = 9.8838x10^7 cm = 988 km

So the radius of a spherical meteorite holding the required amount of water would be approximately 988 km.

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