afiqahiryani1
2022-04-20
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Answered 2022-06-20
Author has **208** answers

To get the conversion factors:

$1000kg=1Mg$

$1{m}^{3}={10}^{6}c{m}^{3}$

$1c{m}^{3}=1mL$

$1mL={10}^{3}\mu L$

Now we can substitute with conversion factors to get

$\rho ={10}^{18}\times \frac{kg}{{m}^{3}}\times \frac{1\times {m}^{3}}{{10}^{6}\times c{m}^{3}}\times \frac{1\times c{m}^{3}}{1\times mL}\times \frac{1\times mL}{{10}^{3}\times \mu L}\times \frac{1\times Mg}{1000\times kg}$

cancel similar units to het that

$\rho ={10}^{6}\frac{Mg}{\mu L}$

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