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Answer :
Sure! Let's solve the problem step-by-step using the information provided.
### Part 1: Calculating the Density of the First Object
1. Identify the given values:
- Mass of the object = [tex]\( 3.00 \times 10^2 \)[/tex] grams (or 300 grams)
- Volume of the object = 97.7 mL
2. Recall the formula for density:
[tex]\[
\text{Density} = \frac{\text{Mass}}{\text{Volume}}
\][/tex]
3. Substitute the given values into the formula:
[tex]\[
\text{Density} = \frac{300 \, \text{g}}{97.7 \, \text{mL}}
\][/tex]
4. Calculate the density:
- The calculated density is approximately 3.07 [tex]\(\frac{\text{g}}{\text{mL}}\)[/tex].
5. Consider significant figures:
- The mass (300 g) has 3 significant figures, and the volume (97.7 mL) has 3 significant figures. Thus, the density should also be expressed with 3 significant figures: 3.07 [tex]\(\frac{\text{g}}{\text{mL}}\)[/tex].
### Part 2: Calculating the Density of the Isopropyl Alcohol-Water Mixture
1. Identify the given values:
- Mass of the mixture = 72.5 grams
- Volume of the mixture = 88.9 mL
2. Use the same density formula:
[tex]\[
\text{Density} = \frac{\text{Mass}}{\text{Volume}}
\][/tex]
3. Substitute the given values into the formula:
[tex]\[
\text{Density} = \frac{72.5 \, \text{g}}{88.9 \, \text{mL}}
\][/tex]
4. Calculate the density:
- The calculated density is approximately 0.816 [tex]\(\frac{\text{g}}{\text{mL}}\)[/tex].
5. Consider significant figures:
- The mass (72.5 g) has 3 significant figures, and the volume (88.9 mL) also has 3 significant figures. Therefore, the density should be expressed with 3 significant figures: 0.816 [tex]\(\frac{\text{g}}{\text{mL}}\)[/tex].
In conclusion, the density of the first object is 3.07 [tex]\(\frac{\text{g}}{\text{mL}}\)[/tex], and the density of the isopropyl alcohol-water mixture is 0.816 [tex]\(\frac{\text{g}}{\text{mL}}\)[/tex].
### Part 1: Calculating the Density of the First Object
1. Identify the given values:
- Mass of the object = [tex]\( 3.00 \times 10^2 \)[/tex] grams (or 300 grams)
- Volume of the object = 97.7 mL
2. Recall the formula for density:
[tex]\[
\text{Density} = \frac{\text{Mass}}{\text{Volume}}
\][/tex]
3. Substitute the given values into the formula:
[tex]\[
\text{Density} = \frac{300 \, \text{g}}{97.7 \, \text{mL}}
\][/tex]
4. Calculate the density:
- The calculated density is approximately 3.07 [tex]\(\frac{\text{g}}{\text{mL}}\)[/tex].
5. Consider significant figures:
- The mass (300 g) has 3 significant figures, and the volume (97.7 mL) has 3 significant figures. Thus, the density should also be expressed with 3 significant figures: 3.07 [tex]\(\frac{\text{g}}{\text{mL}}\)[/tex].
### Part 2: Calculating the Density of the Isopropyl Alcohol-Water Mixture
1. Identify the given values:
- Mass of the mixture = 72.5 grams
- Volume of the mixture = 88.9 mL
2. Use the same density formula:
[tex]\[
\text{Density} = \frac{\text{Mass}}{\text{Volume}}
\][/tex]
3. Substitute the given values into the formula:
[tex]\[
\text{Density} = \frac{72.5 \, \text{g}}{88.9 \, \text{mL}}
\][/tex]
4. Calculate the density:
- The calculated density is approximately 0.816 [tex]\(\frac{\text{g}}{\text{mL}}\)[/tex].
5. Consider significant figures:
- The mass (72.5 g) has 3 significant figures, and the volume (88.9 mL) also has 3 significant figures. Therefore, the density should be expressed with 3 significant figures: 0.816 [tex]\(\frac{\text{g}}{\text{mL}}\)[/tex].
In conclusion, the density of the first object is 3.07 [tex]\(\frac{\text{g}}{\text{mL}}\)[/tex], and the density of the isopropyl alcohol-water mixture is 0.816 [tex]\(\frac{\text{g}}{\text{mL}}\)[/tex].
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