View Full Version : problem with density
starsun
Sep11-04, 04:27 PM
does anyone know how to solve this problem? ive been trying to figure it out for so long but obviously i cant. heres the problem:
Density of seawater is 1.1g/cm cubed. Determine the mass of 10mile cubed of seawater.
thanks :smile:
gravenewworld
Sep11-04, 05:00 PM
i can show you the units but you will have to look up the conversion factors.
1.1 g seawater/cm^3 x 100^3 cm^3/1m^3 x (how ever many meters in 1 mile)^3 m^3/mi^3 x 10 mi^3= x g of seawater.
Azrioch
Sep11-04, 06:12 PM
Well, density is Mass divided by Volume
d = m/v
d(v) = m
So, you only need to convert 10 miles cubed into centimeters cubed.
Just look up on a simple conversion table.
Mile to Cm conversion.. hmm.. *pulls out chemistry text book*
Well, I've only got miles to km here, which is 1 mi = 1.609 km.
Soo... that means 10(1.609x10^5) will give you your volume.
10^5 because kilo--> centi has a five place difference.
Kilo is 1000 and centi is .01
So that means there is 10^5 centimeters in one kilometer.
Then just divide.
I think that's correct.
We're doing this same thing in my General Chemistry I course.
density should be in g/cm^3, the latter corresponds to mass/volume. You need to convert a cube having a volume of 10 x 10 x 10 miles, to cm^3...its standard factor label calculations.
Gokul43201
Sep12-04, 12:18 AM
Mile to Cm conversion.. hmm.. *pulls out chemistry text book*
Well, I've only got miles to km here, which is 1 mi = 1.609 km.
Soo... that means 10(1.609x10^5) will give you your volume.
I'm sure Azrioch meant to write 10(1.609x10^5)^3, which would be the correct conversion of 10 cubic miles into cubic centimeters.
Just to make sure there's no misunderstanding arising out of convention, this is
10 * (1.609*{10}^5)^3
Azrioch
Sep12-04, 07:30 AM
Ah, yes, whoops...
:uhh:
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