Having real trouble with this question...

fawd · #1 $Old$ November 1st, 2009, 05:27 PM

Hello everyone... I am in a university-level science coarse and I am having real trouble with this question on our HW assignment as I am not a math oriented individual.... Any help is greatly appreciated. I know its long, but I am really terrible at this whole math thing.

Thanks again..

Their starting point was the assumption that the total mass or iridium in the K/T
layer was equal to the mass of iridium in the original impactor:

m
Ir(layer) = mIr(impactor)

1) First you must estimate the mass of iridium in the K/T layer. Use the following
information:
a. On average, the iridium layer is

H = 3 cm thick (what is it in SI units?).
b. The density of the K/T layer containing iridium is d = 2.5 g/cm3 (what is it in SI
units?).
c. On average, the layer had a concentration of iridium of CIr(layer) =20 parts per
billion (20 ppb) by weight. (Hint: in scientific notation, 1 billion is 109, so 1 ppb is
equivalent to 1/109 or 10-9)
d. Assume that the iridium concentration was uniform around the Earth.
e. The total area of the Earth is AEarth = 4 π×(REarth)2

where the radius of Earth is

REarth = 6378 km (what is it in SI units?).
f. The total mass of the K/T layer, m(layer), is given by the area of the Earth (AEarth)
multiplied by the layer thickness (H) multiplied by the layer density (d).
g. The total mass of iridium in the K/T layer is given by the mass of the layer
multiplied by the iridium concentration:

m
Ir(layer) = m(layer)×CIr(layer)

CaptainBlack · #2 $Old$ November 1st, 2009, 11:29 PM

Quote:

Originally Posted by fawd $View Post$

Hello everyone... I am in a university-level science coarse and I am having real trouble with this question on our HW assignment as I am not a math oriented individual.... Any help is greatly appreciated. I know its long, but I am really terrible at this whole math thing.

Thanks again..

Their starting point was the assumption that the total mass or iridium in the K/T
layer was equal to the mass of iridium in the original impactor:

m

Ir(layer) = mIr(impactor)

1) First you must estimate the mass of iridium in the K/T layer. Use the following

information:
a. On average, the iridium layer is H = 3 cm thick (what is it in SI units?).
b. The density of the K/T layer containing iridium is d = 2.5 g/cm3 (what is it in SI
units?).
c. On average, the layer had a concentration of iridium of CIr(layer) =20 parts per
billion (20 ppb) by weight. (Hint: in scientific notation, 1 billion is 109, so 1 ppb is
equivalent to 1/109 or 10-9)
d. Assume that the iridium concentration was uniform around the Earth.
e. The total area of the Earth is AEarth = 4 π×(REarth)2
where the radius of Earth is

REarth = 6378 km (what is it in SI units?).

f. The total mass of the K/T layer, m(layer), is given by the area of the Earth (AEarth)

multiplied by the layer thickness (H) multiplied by the layer density (d).
g. The total mass of iridium in the K/T layer is given by the mass of the layer
multiplied by the iridium concentration:
m

Ir(layer) = m(layer)×CIr(layer)

First calculate the mass if the iridium layer. This is $A\times d \times \rho$ where $A$ is the surface area of the Earth and $d$ is the thickness of the iridium layer and $\rho$ the density of the layer. All of these should be in SI units, that is $A$ is in square metres, $d$ in metres, and $\rho$ in kg/cubic metre.

$A=4 \pi R^2$

where $R$ is the radius of the Earth (in metres).

CB