Scenes AD represent atomic-scale views of different samples of substances: (a) Under one set of conditions, the substances in A and B mix and the result is depicted in C. Does this represent a chemical or a physical change? (b) Under a second set of conditions, the same substances mix and the result is depicted in D. Does this represent a chemical or a physical change? (c) Under a third set of conditions, the sample depicted in C changes to that in D. Does this represent a chemical or a physical change? (d) When the change in part (c) occurs, does the sample have different chemical properties? Physical properties?
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Textbook Solutions for Chemistry: The Molecular Nature of Matter and Change
Question
Asbestos is a fibrous silicate mineral with remarkably high tensile strength. But it is no longer used because airborne asbestos particles can cause lung cancer. Grunerite, a type of asbestos, has a tensile strength of \(3.510^{2} \mathrm{~kg} / \mathrm{mm}^{2}\) (thus, a strand of grunerite with a \(1-\mathrm{mm}^{2}\) cross-sectional area can hold up to \(3.510^{2} \mathrm{~kg}\)). The tensile strengths of aluminum and Steel No. 5137 are \(2.510^{4} \mathrm{lb} / \mathrm{in}^{2}\) and \(5.010^{4} \mathrm{lb} / \mathrm{in}^{2}\), respectively. Calculate the cross-sectional area \(\left(\text { in } \mathrm{mm}^{2}\right)\) of wires of aluminum and of Steel No. 5137 that have the same tensile strength as a fiber of grunerite with a cross-sectional area of \(1.0 \mathrm{~m}^{2}\).
Solution
Step 1 of 5
From the given,
The tensile strength of steel = \(5.0 \times 10^{4} \mathrm{lb} / \mathrm{in}^{2}\)
Convert the tensile strength \(l b / \text { in }^{2}\) into \(\mathrm{kg} / \mathrm{mm}^{2}\).
1 kg = 2.205 lb
39.37 in = 1000 mm
\(\begin{aligned}
\text { Strength } & =\frac{5.0 \times 10^{4} \mathrm{lb}}{\mathrm{in}^{2}} \times \frac{1 \mathrm{~kg}}{2.205 \mathrm{lb}} \times\left(\frac{0.03937}{\mathrm{~mm}}\right)^{2} \\
& =\frac{5.0 \times 10^{4} \mathrm{lb}}{2.505} \times\left(\frac{0.03937}{\mathrm{~mm}}\right)^{2} \\
& =1.996 \times 10^{4} \times \frac{0.00155}{\mathrm{~mm}^{2}} \\
& =3.093 \mathrm{~kg} / \mathrm{mm}^{2}
\end{aligned}\)
The strength of steel is \(3.093 \mathrm{~kg} / \mathrm{mm}^{2}\).
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