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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 student's answer.

[tex]\[
\frac{(0.38 \, L)\left(\frac{1 \, mL}{10^{-3} \, L}\right)\left(2.00 \frac{ g }{ mL }\right)}{\left(73.34 \frac{ g }{ mol }\right)}=?
\][/tex]

[tex]\(\square\)[/tex]



Answer :

GinnyAnswer
To determine the units of the student's final answer in the given stoichiometry problem, let's carefully analyze and cancel out the units step by step.

The equation provided is:
[tex]$ \frac{(0.38\ L)\left(\frac{1\ mL}{10^{-3}\ L}\right)\left(2.00\ \frac{g}{mL}\right)}{\left(73.34\ \frac{g}{mol}\right)} = ? $[/tex]

### Step-by-Step Unit Analysis:

1. Convert volume from liters to milliliters:
[tex]$ 0.38\ L \times \left(\frac{1\ mL}{10^{-3} L}\right) $[/tex]
- Units:
[tex]$ L \times \left(\frac{mL}{L}\right) $[/tex]
- Result:
[tex]$ mL $[/tex]

2. Calculate the mass in grams using the density:
[tex]$ (0.38\ L \times 1000 \frac{mL}{L}) \times 2.00\ \frac{g}{mL} $[/tex]
- Units:
[tex]\[ \left(mL\right) \times \left(\frac{g}{mL}\right) \][/tex]
- Result:
[tex]\[ g \][/tex]

3. Calculate the amount in moles by dividing by the molar mass:
[tex]$ \frac{(0.38\ L \times 1000 \frac{mL}{L}) \times 2.00\ \frac{g}{mL}}{73.34\ \frac{g}{mol}} $[/tex]
- Units:
[tex]\[ \frac{g}{\frac{g}{mol}} = mol \][/tex]

### Conclusion:
After canceling out all the intermediate units, the final unit of the student's answer is:
[tex]$ \boxed{mol} $[/tex]
nagabhupendragurram

Answer:

d

Explanation:

13

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