[ryneclipses]

Ok so i have this homework problem that I have been trying to dissect but I just cant seem to get the correct answer.  Its not graded so I'm not trying to just hassle you guys for free answers, I just want to understand this basic problem so I can move on!  Here is the problem:
"a fertilizer suggests an application of 2.06 X 10^-1 kg m^-2     convert to lbs foot^-2
i know that 1 kg = 2.2 lbs
and 1 m = 3.28 feet
how do i setup an equation for this? thanks

[ryneclipses]

I did attempt to answer this by moving the decimal to .206 to make .206kgm^-2 then multiplying the kg and m conversion .206 (2.2) (3.28) and my answer was 1.47 but this was incorrect

[Arkcon]

Try to set up a dimensional analysis:  http://www.brightstorm.com/science/chemistry/introduction-to-chemistry/dimensional-analysis/  Express the known value as a fraction, and multiply conversion factors until all the unwanted units cancel, and only the ones you want are left.

[ryneclipses]

I watched the video and read the transcript on dimensional analysis.  I watched a couple more videos on multi-step dimensional analysis.  I set up an equation for my problem as follows:  (2.06 kg X 10^-1 / m^-2) X (2.2 lbs / kg) X (m^-2 / 3.28 ft^-2)  now I am not sure how to multiply these out because of the powers, but I think it should cancel the unwanted units of kg and m^-2.  Any comments on my setup and/or suggestions for completing?

[magician4]

hint:

if I had to recalculate 1 kg/m² to lbs/foot²  (with the known values 1 kg = 2.2 lbs and 1 m = 3.28 feet) I'd start like this:

[tex]1 \ \frac {kg}{(m)^2} \ =   1    \frac {2.2 \ lbs }{(3.28 \ foot )^2} \ =   \frac {1 \cdot \ 2.2}{(3.28)^2}  \cdot  \frac {lbs}{foot^2}[/tex]

regards

Ingo


p.s.: sooner or later the ongoing  use of these antique units will kill more than just an expensive Mars mission ...