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Sagot :
Sure, let's carefully work through the problem step-by-step to identify the metal sample given to Abner by calculating its density.
### Step-by-Step Solution:
1. Gather Information:
- Mass of the metal sample: [tex]\( m = 93.6 \)[/tex] grams
- Density values from the table:
[tex]\[ \begin{aligned} \text{Aluminum} & : 2.7 \, \text{g/cm}^3 \\ \text{Magnesium} & : 1.7 \, \text{g/cm}^3 \\ \text{Copper} & : 8.9 \, \text{g/cm}^3 \\ \text{Steel} & : 7.9 \, \text{g/cm}^3 \\ \text{Iron} & : 7.8 \, \text{g/cm}^3 \\ \text{Tin} & : 7.3 \, \text{g/cm}^3 \\ \text{Lead} & : 11.3 \, \text{g/cm}^3 \\ \text{Zinc} & : 7.2 \, \text{g/cm}^3 \\ \end{aligned} \][/tex]
2. Identify the Volume:
To identify the volume [tex]\( V \)[/tex] of the metal sample, we need the dimensions. In the absence of dimensional measurements, we can work backwards using density values. Assume the volume [tex]\( V \)[/tex] is correctly measured to identify the metal.
Given:
[tex]\[ \rho = \frac{m}{V} \][/tex]
3. Calculation of Volume for Each Metal using Mass:
We'll compute the hypothetical volume for each metal using given densities:
[tex]\[ V_{\text{metal}} = \frac{m}{\text{density}} \][/tex]
[tex]\[ \begin{aligned} V_{\text{Aluminum}} & = \frac{93.6}{2.7} \approx 34.67 \, \text{cm}^3 \\ V_{\text{Magnesium}} & = \frac{93.6}{1.7} \approx 55.06 \, \text{cm}^3 \\ V_{\text{Copper}} & = \frac{93.6}{8.9} \approx 10.52 \\, \text{cm}^3 \\ V_{\text{Steel}} & = \frac{93.6}{7.9} \approx 11.85 \, \text{cm}^3 \\ V_{\text{Iron}} & = \frac{93.6}{7.8} \approx 12.00 \, \text{cm}^3 \\ V_{\text{Tin}} & = \frac{93.6}{7.3} \approx 12.82 \, \text{cm}^3 \\ V_{\text{Lead}} & = \frac{93.6}{11.3} \approx 8.29 \, \text{cm}^3 \\ V_{\text{Zinc}} & = \frac{93.6}{7.2} \approx 13.00 \, \text{cm}^3 \\ \end{aligned} \][/tex]
4. Comparison and Conclusion:
By comparing the calculated volumes, the volume closest to a typical solid measurement is [tex]\( 12 \, \text{cm}^3 \)[/tex].
The calculated volume for Iron is [tex]\( 12.00 \, \text{cm}^3 \)[/tex], matching a realistic volume for a 93.6-gram sample.
### Justification:
The metal sample given to Abner is most likely Iron because the calculated volume of the sample using the density of Iron (7.8 g/cm³) is approximately [tex]\( 12.00 \, \text{cm}^3 \)[/tex], which is a plausible physical volume for such a sample. This aligns with the given mass of 93.6 grams for iron, strongly suggesting that the metal sample is indeed Iron.
### Step-by-Step Solution:
1. Gather Information:
- Mass of the metal sample: [tex]\( m = 93.6 \)[/tex] grams
- Density values from the table:
[tex]\[ \begin{aligned} \text{Aluminum} & : 2.7 \, \text{g/cm}^3 \\ \text{Magnesium} & : 1.7 \, \text{g/cm}^3 \\ \text{Copper} & : 8.9 \, \text{g/cm}^3 \\ \text{Steel} & : 7.9 \, \text{g/cm}^3 \\ \text{Iron} & : 7.8 \, \text{g/cm}^3 \\ \text{Tin} & : 7.3 \, \text{g/cm}^3 \\ \text{Lead} & : 11.3 \, \text{g/cm}^3 \\ \text{Zinc} & : 7.2 \, \text{g/cm}^3 \\ \end{aligned} \][/tex]
2. Identify the Volume:
To identify the volume [tex]\( V \)[/tex] of the metal sample, we need the dimensions. In the absence of dimensional measurements, we can work backwards using density values. Assume the volume [tex]\( V \)[/tex] is correctly measured to identify the metal.
Given:
[tex]\[ \rho = \frac{m}{V} \][/tex]
3. Calculation of Volume for Each Metal using Mass:
We'll compute the hypothetical volume for each metal using given densities:
[tex]\[ V_{\text{metal}} = \frac{m}{\text{density}} \][/tex]
[tex]\[ \begin{aligned} V_{\text{Aluminum}} & = \frac{93.6}{2.7} \approx 34.67 \, \text{cm}^3 \\ V_{\text{Magnesium}} & = \frac{93.6}{1.7} \approx 55.06 \, \text{cm}^3 \\ V_{\text{Copper}} & = \frac{93.6}{8.9} \approx 10.52 \\, \text{cm}^3 \\ V_{\text{Steel}} & = \frac{93.6}{7.9} \approx 11.85 \, \text{cm}^3 \\ V_{\text{Iron}} & = \frac{93.6}{7.8} \approx 12.00 \, \text{cm}^3 \\ V_{\text{Tin}} & = \frac{93.6}{7.3} \approx 12.82 \, \text{cm}^3 \\ V_{\text{Lead}} & = \frac{93.6}{11.3} \approx 8.29 \, \text{cm}^3 \\ V_{\text{Zinc}} & = \frac{93.6}{7.2} \approx 13.00 \, \text{cm}^3 \\ \end{aligned} \][/tex]
4. Comparison and Conclusion:
By comparing the calculated volumes, the volume closest to a typical solid measurement is [tex]\( 12 \, \text{cm}^3 \)[/tex].
The calculated volume for Iron is [tex]\( 12.00 \, \text{cm}^3 \)[/tex], matching a realistic volume for a 93.6-gram sample.
### Justification:
The metal sample given to Abner is most likely Iron because the calculated volume of the sample using the density of Iron (7.8 g/cm³) is approximately [tex]\( 12.00 \, \text{cm}^3 \)[/tex], which is a plausible physical volume for such a sample. This aligns with the given mass of 93.6 grams for iron, strongly suggesting that the metal sample is indeed Iron.
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