Question

Problem with Conversion Factors

A utility worker notices a leak in a water pipe and measures it be 1.9 x 10^2 mL/s. Convert this into L/day (Report the answer in scientific notation and have the correct number of significant figures in your answer too.)

Answer 1

I'm not good in significant figures, but I can try

Answer 2

\[\frac{190mL}{1 second} \times \frac{1L}{1000mL} \times \frac{86400seconds}{1 Day}\]

Answer 3

1L = 1000 mL

1 day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds
1 day = 24 x 60 x 60 = 86400 seconds


1.9 x 10^2 mL/ s --> 2 significant figures

1.9 x 10^2 mL/ s = 1.9 x 10^2 : (1000/86400) L/ day
1.9 x 10^2 mL/ s = 1.9 x 10^2 mL/ s x (86400/1000) L/ day
1.9 x 10^2 mL/ s = 16416 L/ day
1.9 x 10^2 mL/ s = 1,6 x 10^4 L/ day (significant figures)

Answer 4

I mean
1,6 x 10^4 L/ day (2 significant figures)

Answer 5

Thanks :)

Answer 6

you're welcome :)