Let's proceed step-by-step to solve the problem based on the information provided.
### Step 1: Convert Volume to Cubic Centimeters
First, we need to convert the volume of the unknown liquid from liters to cubic centimeters.
Given:
- Volume of the liquid [tex]\(\text{V}_{\text{liters}} = 0.982 \text{ L}\)[/tex]
Since 1 liter is equivalent to 1000 cubic centimeters, we can convert the volume as follows:
[tex]\[
\text{V}_{\text{cm}} = 0.982 \text{ L} \times 1000 = 982.0 \text{ cm}^3
\][/tex]
### Step 2: Calculate the Density
Next, we calculate the density of the liquid using the formula for density:
[tex]\[
\text{Density} = \frac{\text{Mass}}{\text{Volume}}
\][/tex]
Given:
- Mass of the liquid [tex]\(\text{M} = 1.00 \times 10^3 \text{ g}\)[/tex]
- Volume of the liquid [tex]\(\text{V}_{\text{cm}} = 982.0 \text{ cm}^3\)[/tex]
Substitute these values into the formula:
[tex]\[
\text{Density} = \frac{1.00 \times 10^3 \text{ g}}{982.0 \text{ cm}^3} \approx 1.018 \frac{\text{g}}{\text{cm}^3}
\][/tex]
Rounding the density to 3 significant digits, we get:
[tex]\[
\text{Density} = 1.018 \frac{\text{g}}{\text{cm}^3}
\][/tex]
### Step 3: Compare the Calculated Density with Given Densities
We now compare the calculated density with the densities of the known liquids:
[tex]\[
\begin{array}{|c|c|}
\hline
\text{Liquid} & \text{Density} (\frac{\text{g}}{\text{cm}^3}) \\
\hline
\text{Diethylamine} & 0.71 \\
\hline
\text{Ethanolamine} & 1.0 \\
\hline
\text{Carbon Tetrachloride} & 1.6 \\
\hline
\text{Acetone} & 0.79 \\
\hline
\text{Glycerol} & 1.3 \\
\hline
\end{array}
\][/tex]
The calculated density is 1.018 [tex]\(\frac{\text{g}}{\text{cm}^3}\)[/tex]. Comparing this with the known densities, we see that none of the given liquids have a density close to 1.018 [tex]\(\frac{\text{g}}{\text{cm}^3}\)[/tex].
### Conclusion
1. Calculate the density of the liquid. Round your answer to 3 significant digits.
[tex]\[
\text{Density} = 1.018 \frac{\text{g}}{\text{cm}^3}
\][/tex]
2. Given the data above, is it possible to identify the liquid?
[tex]\[
\text{no}
\][/tex]
3. If it is possible to identify the liquid, do so.
Since it's not possible to accurately match the density of 1.018 [tex]\(\frac{\text{g}}{\text{cm}^3}\)[/tex] with any of the given liquids, we cannot identify the liquid.
So, our answers to the questions are:
[tex]\[
\begin{array}{|c|c|}
\hline
\text{Calculate the density of the liquid (rounded to 3 significant digits)} & 1.018 \frac{\text{g}}{\text{cm}^3} \\
\hline
\begin{array}{l}
\text{Given the data above, is it possible to} \\
\text{identify the liquid?}
\end{array} & \text{no} \\
\hline
\end{array}
\][/tex]
