a student sets up the following equation to solve a problem in solution stoichiometry. (the ? stands for a…

a student sets up the following equation to solve a problem in solution stoichiometry. (the ? stands for a number the student is going to calculate.) enter the units of the students answer. $\frac{(30.0 g)(\frac{1 ml}{10^{-3} l})}{(61.06\frac{g}{mol})(7.9\frac{mol}{l})}=?square$

a student sets up the following equation to solve a problem in solution stoichiometry. (the ? stands for a number the student is going to calculate.) enter the units of the students answer. $\frac{(30.0 g)(\frac{1 ml}{10^{-3} l})}{(61.06\frac{g}{mol})(7.9\frac{mol}{l})}=?square$

Answer

Explanation:

Step1: Analyze units in numerator

The numerator has units of (g\times\frac{mL}{10^{- 3}L}). Since (1\ mL = 10^{-3}L), the units of the numerator simplify to (g\times1 = g).

Step2: Analyze units in denominator

The denominator has units of (\frac{g}{mol}\times\frac{mol}{L}= \frac{g}{L}).

Step3: Calculate final units

When we divide the numerator by the denominator (\frac{g}{\frac{g}{L}}), using the rule of dividing by a fraction ((a\div\frac{b}{c}=a\times\frac{c}{b})), we get (L).

Answer:

(L)