Question

Which of the following is a possible equivalence that can be used in a conversion?

ten cubed kL over one L

ten squared daL over one L

ten to the negative second power L over one cL

ten to the negative first power L over one mL

Answer 1

Let me look this one up

Answer 2

the third one

Answer 3

agree with @Claflamme3

Answer 4

Hi! You have your answer, which I agree with, but I want to explain how you'd know that!

Firstly, when you convert to a different unit, the value - what the number represents - doesn't change. You're actually multiplying the number by \(1\). Its a fraction, where the top and bottom are equal. I'll derive a simple conversion that I hope you agree with.

Even if you're in the stubborn U.S., it's important to know the metric system. Like, 1 meter is one hundred centimters. In math, \(1\ [m]=100\ [cm]\)
This would imply that, by dividing by 1 meter,\[\frac{1\ [m]}{1\ [m]}=\frac{100\ [cm]}{1\ [m]}\]\[\qquad\qquad\Downarrow\]\[\frac{\cancel{1\ [m]}}{\cancel{1\ [m]}}=\frac{100\ [cm]}{1\ [m]}\]\[\qquad\qquad\Downarrow\]\[1=\frac{100\ [cm]}{1\ [m]}\]
That sort of thing is always good to know for unit conversions. And reversing that on your multiple choices will check to see if they are good!

\[\frac{10^3\ [kL]}{[L]}\qquad\rightarrow\qquad 10^3\ [kL]\overset{?}{=}[L]\]
\[\frac{10^2\ [daL]}{[L]}\qquad\rightarrow\qquad 10^2\ [daL]\overset{?}{=}[L]\]
\[\frac{10^{-2}\ [L]}{[cL]}\qquad\rightarrow\qquad 10^{-2}\ [L]\overset{?}{=}[cL]\]
\[\frac{10^{-1}\ [L]}{[mL]}\qquad\rightarrow\qquad 10^{-1}\ [L]\overset{?}{=}[mL]\]

And note that \(\Large 10^{-1}=\frac{1}{10^1}\) and that \(\Large 10^{-2}=\frac{1}{10^2}\), just by the meaning of negative exponents. Just for my own fun...

\[\frac{10^3\ [kL]}{[L]}\qquad\rightarrow\qquad 10^3\ [kL]\cancel{=}[L]\]
\[\frac{10^2\ [daL]}{[L]}\qquad\rightarrow\qquad 10^2\ [daL]\cancel{=}[L]\]
\[\frac{10^{-2}\ [L]}{[cL]}\qquad\rightarrow\qquad 10^{-2}\ [L]{=}[cL]\qquad\huge \checkmark \]
\[\frac{10^{-1}\ [L]}{[mL]}\qquad\rightarrow\qquad 10^{-1}\ [L]\cancel{=}[mL]\]