## jakethekicker 2 years ago 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

1. Claflamme3

Let me look this one up

2. Claflamme3

the third one

3. mr.singh

agree with @Claflamme3

4. theEric

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