Answered

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]\[ \left(350 \, \frac{mg}{dL}\right) \left(\frac{10^{-3} \, g}{1 \, mg}\right) \left(\frac{1 \, dL}{10^{-1} \, L}\right) \left(\frac{10^{-3} \, L}{1 \, mL}\right) = \text{ ? } \][/tex]



Answer :

GinnyAnswer
To determine the units of the student's answer, we will carefully analyze and simplify the given equation step by step.

Given the equation:

[tex]\[ \left(350 \ \frac{ \text{mg} }{ \text{dL} }\right) \left(\frac{10^{-3} \ \text{g}}{1 \ \text{mg}}\right) \left(\frac{1 \ \text{dL}}{10^{-1} \ \text{L}}\right) \left(\frac{10^{-3} \ \text{L}}{1 \ \text{mL} }\right) = \text{ ? } \][/tex]

Let's analyze the units in each term and simplify them systematically.

1. First term: [tex]\(350 \ \frac{\text{mg}}{\text{dL}}\)[/tex]
- [tex]\(350\)[/tex] mg per deciliter.

2. Second term: [tex]\(\frac{10^{-3} \ \text{g}}{1 \ \text{mg}}\)[/tex]
- This term converts milligrams to grams.
- [tex]\(10^{-3} \ g\)[/tex] per [tex]\(1 \ \text{mg}\)[/tex].

3. Third term: [tex]\(\frac{1 \ \text{dL}}{10^{-1} \ \text{L}}\)[/tex]
- This term converts deciliters to liters.
- [tex]\(1 \ \text{dL}\)[/tex] per [tex]\(10^{-1} \ \text{L}\)[/tex].

4. Fourth term: [tex]\(\frac{10^{-3} \ \text{L}}{1 \ \text{mL}}\)[/tex]
- This term converts liters to milliliters.
- [tex]\(10^{-3} \ L\)[/tex] per [tex]\(1 \ \text{mL}\)[/tex].

Now we'll write out the equation focusing only on the units:

[tex]\[ \left( \frac{\text{mg}}{\text{dL}} \right) \left( \frac{\text{g}}{\text{mg}} \right) \left( \frac{\text{dL}}{\text{L}} \right) \left( \frac{\text{L}}{\text{mL}} \right) \][/tex]

Next, cancel out the appropriate units step-by-step:

1. The unit 'mg' in the first term ([tex]\(\frac{\text{mg}}{\text{dL}}\)[/tex]) cancels with the unit 'mg' in the second term ([tex]\(\frac{\text{g}}{\text{mg}}\)[/tex]).
2. The unit 'dL' in the first combined term ([tex]\(\frac{\text{dL}}{\text{L}}\)[/tex]) cancels with the unit 'dL' in the third term ([tex]\(\frac{\text{dL}}{\text{L}}\)[/tex]).
3. The unit 'L' in the first combined term ([tex]\(\frac{\text{g}}{\text{L}}\)[/tex] after the previous cancellations) cancels with the unit 'L' in the fourth term ([tex]\(\frac{\text{L}}{\text{mL}}\)[/tex]).

After all these cancellations, we are left with:

[tex]\[ \frac{\text{g}}{\text{mL}} \][/tex]

Therefore, the units of the student's answer simplify to:

[tex]\[ \boxed{\frac{\text{g}}{\text{mL}}} \][/tex]

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