Find the best answers to your questions at Westonci.ca, where experts and enthusiasts provide accurate, reliable information. Explore thousands of questions and answers from a knowledgeable community of experts on our user-friendly platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
The unit cell volume of the crystal is [tex]V_c = 9.9084*10^-^2^3 cm^3 / unit cell[/tex] and the density of titanium is calculated as 4.71g/cm^3 while the literature value is 4.5 g/cm^3
Data;
- radius = 0.1445 nm
- c/a = 1.58
- A = 46.88 g/mol
Unit Cell Volume
The unit cell volume can be calculated as
[tex]V_c = 6R^2c\sqrt{3}\\[/tex]
let's substitute the values into the formula
[tex]V_c = 6 * (1.445*10^-^8)^2 * 1.58 8 * a * \sqrt{3} \\a = 2R\\V_c = 6 * (1.445*10^-^8)^2 * 1.58* 2 * 1.445*10^-^8 * \sqrt{3}\\ V_c = 9.984*10^-^2^3 cm^3 / unit cell[/tex]
The unit cell volume of the crystal is [tex]V_c = 9.9084*10^-^2^3 cm^3 / unit cell[/tex]
Density of Ti
The density of titanium can be calculated as
[tex]\rho = \frac{nA}{V_c N_a}[/tex]
- n = 6 for hcp
- A = 46.88 g/mol
- Na = Avogadro's number
let's substitute the values into the formula
[tex]\rho = \frac{nA}{V_c Na}\\ \rho = \frac{6*46.88}{9.9084*10^-^2^3* 6.023*10^2^3} \\\rho = 4.71 g/cm^3[/tex]
The density of titanium is calculated as 4.71g/cm^3 while the literature value is 4.5 g/cm^3
Learn more on crystal lattice here;
https://brainly.com/question/6610542
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
