# How do you correctly round this problem (30.00 g)/ (10.0 mL)?

Dec 30, 2017

$\frac{3 g}{m L}$

#### Explanation:

You can just divide to get $\frac{3 g}{m L}$. Is this talking about the density of a substance?

Dec 30, 2017

Please see the step process below...

#### Explanation:

Firstly, you have to convert the $m L \to L$

Though it depends on what you want your final unit to be..

But in this case, am going to convert it;

$m L \to L$

$1000 m L \to 1 L$

$10 m L \to x L$

$x = \frac{1 L \times 10 m L}{1000 m L}$

x = (1L xx cancel(10mL))/(100 xx cancel(10mL)

$x = \frac{1}{10} L$

$x = 0.01 L$

Hence we have;

$\frac{30.00 g}{10.0 m L} \Rightarrow \frac{30 g}{0.01 L}$

$\Rightarrow 3000.00 g {L}^{-} 1 \mathmr{and} 3.00 g m {L}^{-} 1$

Dec 30, 2017

Well, you are dividing, so you keep the fewest number of significant figures. $\text{10.0 mL}$ has $3$, and $\text{30.00 g}$ has four.

All trailing zeroes after the decimal place are significant, and zeroes sandwiched between a decimal place and a nonzero digit are also significant.

So you just get:

"30.00 g"/"10.0 mL" = color(blue)("3.00 g"/"mL")