To fill in the missing part of the student's equation, let's break down the problem step-by-step using the given answer.
1. Identify the given quantities:
- The mass of the substance: [tex]\( 117 \ \text{g} \)[/tex]
- Volume conversion factor: [tex]\( \frac{1 \ \text{mL}}{10^{-3} \ \text{L}} \)[/tex]
- A placeholder factor in the denominator: [tex]\( 53.40 \ [\square] \)[/tex]
- Molarity of the solution: [tex]\( 1.8 \ \text{mol/L} \)[/tex]
- Target volume: [tex]\( 1.2 \times 10^3 \ \text{mL} \)[/tex]
2. Write down the equation:
[tex]\[
\frac{(117 \ \text{g}) \left( \frac{1 \ \text{mL}}{10^{-3} \ \text{L}} \right)}{(53.40 \ [\square]) \left(1.8 \ \text{mol/L} \right)} = 1.2 \times 10^3 \ \text{mL}
\][/tex]
3. Simplify the given terms:
- The volume conversion factor simplifies to [tex]\( 1 \text{mL} = 1000 \ \text{L} \)[/tex]:
[tex]\[
\frac{1 \ \text{mL}}{10^{-3} \ \text{L}} = 1000 \ \text{mL/L}
\][/tex]
- Therefore, the numerator will be:
[tex]\[
117 \ \text{g} \times 1000 \ = 117000 \ \text{g \cdot mL/L}
\][/tex]
4. Set up the equation to solve for the missing part:
The equation now looks like:
[tex]\[
\frac{117000 \ [g \cdot mL/L]}{(53.40 \ [\square]) \left(1.8 \ \text{mol/L} \right)} = 1.2 \times 10^3 \ \text{mL}
\][/tex]
5. Rearrange to isolate the missing part:
Solving for the missing part, we get:
[tex]\[
53.40 \ [\square] = \frac{117000 \ [g \cdot mL/L]}{1.2 \times 10^3 \ [\text{mL}] \times 1.8 \ [\text{mol/L}]}
\][/tex]
Simplify this:
[tex]\[
53.40 [\square] = \frac{117000}{1.2 \times 10^3 \times 1.8}
\][/tex]
6. Calculate the missing part:
[tex]\[
53.40 [\square] = \frac{117000}{2160} \approx 54.1667
\][/tex]
Therefore, the missing part of the equation is approximately [tex]\( 54.1667 \)[/tex]. The filled-in equation is:
[tex]\[
\frac{(117 \ \text{g}) \left( \frac{1 \ \text{mL}}{10^{-3} \ \text{L}} \right)}{(53.40 \times 54.1667) \left(1.8 \ \text{mol/L} \right)} = 1.2 \times 10^3 \ \text{mL}
\][/tex]
