2
$\begingroup$

I have a problem that says "a small hole is $20.7~\mathrm{cm^2}$ wide. What is the area in $\mathrm{km^2}$."

I started doing it like this:

$20.7~\mathrm{cm^2} \cdot 0.01~\mathrm{m^2} / 1~\mathrm{cm^2} = 0.207$

$0.207 \cdot 1~\mathrm{m^2} / 1000~\mathrm{km^2} = 0.000207$ or $2.07\cdot10^{-4}$

My answer says it should be $10^{-9}$. What am I missing here?

$\endgroup$
  • 2
    $\begingroup$ This question appears to be off-topic because it belongs to Mathematics S.E. $\endgroup$ – ashu Sep 4 '14 at 3:28
  • 3
    $\begingroup$ @ashu Unit transformations are a large component of Chemistry and Chemistry related analyses. $\endgroup$ – LordStryker Sep 4 '14 at 13:55
4
$\begingroup$

1 square centimeter = 0.0001 square meters

$$(20.7\ \mathrm{cm}^2) \left ( \dfrac{0.0001\ \mathrm{m}^2}{1\ \mathrm{cm}^2} \right ) = 2.07 \times 10^{-3}\ \mathrm{m}^2$$

1 square meter = 1 x 10^(-6) square kilometers

$$(2.07\times10^{-3}\ \mathrm{m}^2) \left ( \dfrac{1\times10^{-6}\ \mathrm{km}^2}{1\ \mathrm{m}^2}\right) = 2.07 \times 10^{-9}\ \mathrm{km}^2$$

$\endgroup$
  • $\begingroup$ Just to be I'm getting this right and it's not just a coincidence, is it the normal conversion (like cm to m is -2, and m to km is -3) times the unit's exponent? So since it's squared, it's say cm^2 so -2*2, (4), and -3*2, (-6), for the other? $\endgroup$ – Caesium-133 Sep 3 '14 at 22:04
  • 1
    $\begingroup$ Exactly, you've got it! $\endgroup$ – ron Sep 3 '14 at 22:07
  • $\begingroup$ Ok, so in that one I just had 20.7. if it was 20.7 x 10^(-5), or any exponent in sci. notation form adding in, what am I doing with that part? Adding the combo of the conversion and unit exponents to the one in the sci. notation? Or? $\endgroup$ – Caesium-133 Sep 3 '14 at 22:28
  • $\begingroup$ When you multiply powers of 10, you add the exponents, so 20.7 x 10^(-4) x 10^(12) = 20.7 x 10^8. Is that what you're asking about? $\endgroup$ – ron Sep 3 '14 at 22:39
  • $\begingroup$ Like 1.36 x10^(-6)cm^3 to m^3 $\endgroup$ – Caesium-133 Sep 3 '14 at 22:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.