Solve Density Problem: Mass of 10mi³ Seawater

[starsun]

does anyone know how to solve this problem? I've been trying to figure it out for so long but obviously i cant. here's the problem:

Density of seawater is 1.1g/cm cubed. Determine the mass of 10mile cubed of seawater.

thanks

 

[gravenewworld]

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]

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 textbook*
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.

 

[GCT]

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]

Mile to Cm conversion.. hmm.. *pulls out chemistry textbook*
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]

Ah, yes, whoops...

:uhh: